Parabola

Given that we need to find the equation of the parabola whose focus is S(3, 0) and directrix(M) is 3x + 4y - 1 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 25x² + 25y² - 150x + 225 = 9x² + 16y² - 6x - 8y + 24xy + 1

⇒ 16x² + 9y² - 24xy - 144x + 8y + 224 = 0

∴The equation of the parabola is 16x² + 9y² - 24xy - 144x + 8y + 224 = 0

Given that we need to find the equation of the parabola whose focus is S(1, 1) and directrix(M) is x + y + 1 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 2x² + 2y² - 4x - 4y + 4 = x² + y² + 2x + 2y + 2xy + 1

⇒ x² + y² + 2xy - 6x - 6y + 3 = 0

∴The equation of the parabola is x² + y² + 2xy - 6x - 6y + 3 = 0.

Given that we need to find the equation of the parabola whose focus is S(0, 0) and directrix(M) is 2x - y - 1 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 5x² + 5y² = 4x² + y² - 4x + 2y - 4xy + 1

⇒ x² + 4y² + 4xy + 4x - 2y - 1 = 0

∴The equation of the parabola is x² + 4y² + 4xy + 4x - 2y - 1 = 0.

Given that we need to find the equation of the parabola whose focus is S(2, 3) and directrix(M) is x - 4y + 3 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 17x² + 17y² - 68x - 102y + 221 = x² + 16y² + 6x - 24y - 8xy + 9

⇒ 16x² + y² + 8xy - 74x - 78y + 212 = 0

∴The equation of the parabola is 16x² + y² + 8xy - 74x - 78y + 212 = 0.

Given that we need to find the equation of the parabola whose focus is S(2, 3) and directrix(M) is x - 4y + 3 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 17x² + 17y² - 68x - 102y + 221 = x² + 16y² + 6x - 24y - 8xy + 9

⇒ 16x² + y² + 8xy - 74x - 78y + 212 = 0

∴The equation of the parabola is 16x² + y² + 8xy - 74x - 78y + 212 = 0.

We know that the length of the latus rectum is twice the perpendicular distance from the focus to the directrix.

⇒

⇒

⇒

∴The length of the latus rectum is .

We need to find the equation of the parabola whose focus is (- 6, 6), and the vertex is (- 2, 2).

We know that the line passing through focus and vertex, i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

⇒ The slope of the axis (m₁) =

⇒

⇒ m₁ = - 1

We know that the products of the slopes of the perpendicular lines is - 1.

Let us assume m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒ - 1.m₂ = - 1

⇒ m₂ = 1

Let us find the point on directrix.

⇒

⇒

⇒ x - 6 = - 4 and y + 6 = 4

⇒ x = 2 and y = - 2

The point on directrix is (2, - 2).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁)

⇒ y - (- 2) = 1(x - 2)

⇒ y + 2 = x - 2

⇒ x - y - 4 = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 2x² + 2y² + 24x - 24y + 144 = x² + y² - 8x + 8y - 2xy + 16

⇒ x² + y² + 2xy + 32x - 32y + 128 = 0

∴The equation of the parabola is x² + y² + 2xy + 32x - 32y + 128 = 0.

We need to find the equation of the parabola whose focus is (0, - 3) and the vertex is (0, 0).

We know that the line passing through focus and vertex i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

⇒ Slope of axis (m₁) =

⇒

⇒ m₁ = - ∞

We know that the product of the perpendicular lines is applicable for non - vertical lines.

Here we got the axis to be parallel to the x - axis.

∴The slope of the directrix is equal to the slope of x - axis i.e., 0.

⇒ m₂ = 0

Let us find the point on directrix.

⇒

⇒

⇒ x = 0 and y - 3 = 0

⇒ x = 0 and y = 3

The point on directrix is (0, 3).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁)

⇒ y - 3 = 0(x - 0)

⇒ y - 3 = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ x² + 12y = 0

∴The equation of the parabola is x² + 12y = 0.

We need to find the equation of the parabola whose focus is (0, - 3) and the vertex is (- 1, - 3).

We know that the line passing through focus and vertex i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

⇒ Slope of axis (m₁) =

⇒

⇒ m₁ = 0

We know that the products of the slopes of the perpendicular lines is - 1 for non - vertical lines.

Here the slope of the axis is equal to the slope of the x - axis. So, the slope of directrix is equal to the slope of y - axis i.e., ∞.

⇒ m₂ = ∞

Let us find the point on directrix.

⇒

⇒

⇒ x + 0 = - 2 and y - 3 = - 6

⇒ x = - 2 and y = - 3

The point on directrix is (- 2, - 3).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁)

⇒ y - (- 3) = ∞(x - (- 2))

⇒

⇒ x + 2 = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² + y² + 6y + 9 = x² + 4x + 4

⇒ y² - 4x + 6y + 5 = 0

∴The equation of the parabola is y² - 4x + 6y + 5 = 0.

We need to find the equation of the parabola whose focus is (a, 0) and the vertex is (a’, 0).

We know that the line passing through focus and vertex i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

⇒ Slope of axis (m₁) =

⇒

⇒ m₁ = 0

We know that the products of the slopes of the perpendicular lines is - 1 for non - vertical lines.

Here the slope of the axis is equal to the slope of the x - axis. So, the slope of directrix is equal to the slope of y - axis i.e., ∞.

⇒ m₂ = ∞

Let us find the point on directrix.

⇒

⇒

⇒ x + a = 2a’ and y = 0

⇒ x = 2a’ - a and y = 0

The point on directrix is (2a’ - a, 0).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁)

⇒ y - (0) = ∞(x - (2a’ - a))

⇒

⇒ x + a - 2a’ = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² + y² - 2ax + a² = x² + a² + 4(a’)² + 2ax - 4aa’ - 4a’x

⇒ y² - (4a - 4a’)x + a² - 4(a’)² + 4aa’ = 0

∴The equation of the parabola is y² - (4a - 4a’)x + a² - 4(a’)² + 4aa’ = 0.

We need to find the equation of the parabola whose focus is (0, 0) and vertex is the point of intersection of lines x + y = 1 and x - y = 3.

On solving the lines, we get the vertex (2, - 1).

We know that the line passing through focus and vertex i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

⇒ Slope of axis (m₁) =

⇒

We know that the products of the slopes of the perpendicular lines is - 1.

Let us assume m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒

⇒ m₂ = 2

Let us find the point on directrix.

⇒

⇒

⇒ x = 4 and y = - 2

The point on directrix is (4, - 2).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁)

⇒ y - (- 2) = 2(x - 4)

⇒ y + 2 = 2x - 8

⇒ 2x - y - 10 = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 5x² + 5y² = 4x² + y² - 40x + 20y - 4xy + 100

⇒ x² + 4y² + 4xy + 40x - 20y - 100 = 0

∴The equation of the parabola is x² + 4y² + 4xy + 40x - 20y - 100 = 0.

Given parabola is y² = 8x. Comparing this with standard parabola y² = 4ax we get,

⇒ 4a = 8

⇒ a = 2

⇒ The vertex is (0, 0)

⇒ The focus is (a, 0) = (2, 0)

⇒ The equation of the axis is y = 0.

⇒ The equation of the directrix is x = - a i.e, x = - 2

⇒ The length of the latus rectum is 4a = 8.

Given parabola is 4x² = y.

Converting to standard form we get,

⇒

Comparing with standard parabola x² = 4ay we get,

⇒

⇒

⇒ The vertex is (0, 0)

⇒ The focus is (0, a) =

⇒ The equation of the axis is x = 0

⇒ The equation of the directrix is y = - a i.e, .

⇒ The length of the latus rectum is 4a = .

Given equation of the parabola is y² - 4y - 3x + 1 = 0

⇒ y² - 4y = 3x - 1

⇒ y² - 4y + 4 = 3x + 3

⇒ (y - 2)² = 3(x + 1)

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ 4b = 3

⇒

⇒ The vertex is (c, a) = (- 1, 2)

⇒ The focus is (b + c, a) =

⇒ The equation of the axis is y - a = 0 i.e, y - 2 = 0

⇒ The equation of the directrix is x - c = - b

⇒ Directrix is

⇒ Directrix is

⇒ Directrix is

⇒ Length of latus rectum is 4b = 3.

Given equation of the parabola is y² - 4y + 4x = 0

⇒ y² - 4y = - 4x

⇒ y² - 4y + 4 = - 4x + 4

⇒ (y - 2)² = - 4(x - 1)

Comparing with the standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ 4b = 4

⇒ b = 1

⇒ The vertex is (c, a) = (1, 2)

⇒ The focus is (b + c, a) = (1-1, 2) = (0, 2)

⇒ The equation of the axis is y - a = 0 i.e, y - 2 = 0

⇒ The equation of the directrix is x - c = b

⇒ Directrix is x - 1 = 1

⇒ Directrix is x = 1 + 1

⇒ Directrix is x = 2

⇒ Length of latus rectum is 4b = 4.

Given equation of the parabola is y² + 4x + 4y - 3 = 0

⇒ y² + 4y = - 4x + 3

⇒ y² + 4y + 4 = - 4x + 7

⇒

Comparing with the standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ 4b = 4

⇒ b = 1

⇒ The vertex is (c, a) =

⇒ The focus is (- b + c, a) =

⇒ The equation of the axis is y - a = 0 i.e, y + 2 = 0

⇒ The equation of the directrix is x - c = b

⇒ Directrix is

⇒ Directrix is

⇒ Directrix is

⇒ Length of latus rectum is 4b = 4.

Given equation of the parabola is y² = 8x + 8y

⇒ y² - 8y = 8x

⇒ y² - 8y + 16 = 8x + 16

⇒ (y - 4)² = 8(x + 2)

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ 4b = 8

⇒ b = 2

⇒ The vertex is (c, a) = (- 2, 4)

⇒ The focus is (b + c, a) =

⇒ The equation of the axis is y - a = 0 i.e, y - 4 = 0

⇒ The equation of the directrix is x - c = - b

⇒ Directrix is x - (-2) = -2

⇒ Directrix is x = -2 -2

⇒ Directrix is x = -4

⇒ Length of latus rectum is 4b = 8.

Given equation of the parabola is 4(y - 1)² = - 7(x - 3)

⇒

Comparing with the standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒

⇒

⇒ The vertex is (c, a) = (3, 1)

⇒ The focus is (- b + c, a) =

⇒ The equation of the axis is y - a = 0 i.e, y - 1 = 0

⇒ The equation of the directrix is x - c = b

⇒ Directrix is

⇒ Directrix is

⇒ Directrix is

⇒ Length of latus rectum is 4b .

Given equation of the parabola is y² = 5x - 4y - 9

⇒ y² + 4y = 5x - 9

⇒ y² + 4y + 4 = 5x - 5

⇒ (y + 2)² = 5(x - 1)

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ 4b = 5

⇒

⇒ The vertex is (c, a) = (1, - 2)

⇒ The focus is (b + c, a) =

⇒ The equation of the axis is y - a = 0 i.e, y + 2 = 0

⇒ The equation of the directrix is x - c = - b

⇒ Directrix is

⇒ Directrix is

⇒ Directrix is

⇒ Length of latus rectum is 4b = 5.

Given equation of the parabola is x² + y = 6x - 14

⇒ x² - 6x = - y - 14

⇒ x² - 6y + 9 = - y - 5

⇒ (x - 3)² = - (y + 5)

Comparing with the standard form of parabola (x - a)² = - 4b(y - c) we get,

⇒ 4b = 1

⇒

⇒ The vertex is (a, c) = (3, - 5)

⇒ The focus is (a, - b + c) =

⇒ The equation of the axis is x - a = 0 i.e, x - 3 = 0

⇒ The equation of the directrix is y - c = b

⇒ Directrix is

⇒ The directrix is

⇒ Directrix is

⇒ Length of latus rectum is 4b = 1.

Let AB be the double ordinate of length 8p for the parabola y² = 4px.

Comparing with standard form we get y² = 4ax we get,

⇒ axis is y = 0

⇒ vertex is O(0, 0).

We know that double ordinate is perpendicular to the axis.

Let us assume that the point at which the double ordinate meets the axis is (x₁, 0).

Then the equation of the double ordinate is y = x₁. It meets the parabola at the points (x₁, 4p) and (x₁, - 4p) as its length is 8p.

Let us find the value of x₁ by substituting in the parabola.

⇒ (4p)² = 4p(x₁)

⇒ x₁ = 4p.

The extremities of the double ordinate are A(4p, 4p) and B(4p, - 4p).

Let us find assume the slopes of OA and OB be m₁ and m₂. Let us find their values.

⇒

⇒

⇒ m₁ = 1

⇒

⇒

⇒ m₂ = - 1

⇒ m₁.m₂ = 1. - 1

⇒ m₁.m₂ = - 1

We have got product of slopes - 1. So, the lines OA and OB are perpendicular to each other.

So, the extremities of double ordinate make right angle with vertex.

Given that we need to find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus - rectum.

Comparing it with the standard form of parabola x² = 4by.

⇒ Vertex is 0(0, 0)

⇒ Ends of latus rectum is (2b, b), (- 2b, b)

⇒ 4b = 12

⇒ b = 3

⇒ Ends of latus rectum is (2(3), 3), (- 2(3), 3)

⇒ Ends of latus rectum is A(6, 3), B(- 6, 3)

We know that area of the triangle with the vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

⇒

⇒

⇒

⇒

⇒

⇒ A = 18sq.units.

∴The area of the triangle is 18 sq.units.

Given the equation of directrix is 3x - 4y = 2 and focus is (3, 3).

We know that the directrix and axis are perpendicular to each other. The axis also passes through the focus.

Let us find the slope of the directrix.

We know that the slope of the line ax + by + c = 0 is .

⇒

⇒ .

We know that the products of the slopes of the perpendicular lines (non - vertical) is - 1. Let us assume the slope of axis is m₂.

⇒ m₁.m₂ = - 1

⇒

⇒ .

We know that the equation of the line passing through the point (x₁, y₁) and having slope m is (y - y₁) = m(x - x₁)

⇒

⇒ 3(y - 3) = - 4(x - 3)

⇒ 3y - 9 = - 4x + 12

⇒ 4x + 3y = 21

On solving the lines 4x + 3y = 21 and 3x - 4y = 2, we get the intersection point to be .

We know that the length of latus rectum is equal to the twice of the perpendicular distance between directrix and focus.

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒

⇒

⇒

⇒

⇒ L = 2

∴The length of the latus rectum is 2.

Given that we need to find the point on parabola x² = 9y such that the abscissa is three times the ordinate.

Let us assume the point be (3y₁, y₁). Substituting in the parabola we get,

⇒ (3y₁)² = 9(y₁)

⇒ 9y₁² = 9y₁

⇒ y₁² - y₁ = 0

⇒ y₁(y₁ - 1) = 0

⇒ y₁ = 0 or y₁ - 1 = 0

⇒ y₁ = 0 or y₁ = 1

The points is B(3(1), 1) i.e, (3, 1).

∴The point is (3, 1).

Given that we need to find the equation of the parabola whose vertex is at origin and axis along the x - axis and also passing through (2, 3).

We know that the standard equation of the parabola whose axis is x - axis and vertex at origin is y² = 4ax.

Substituting (2, 3) in the equation of parabola.

⇒ (3)² = 4a(2)

⇒ 9 = 8a

⇒

Now,

⇒

⇒

⇒ 2y² = 9x.

∴The equation of the parabola is 2y² = 9x.

Given that we need to find the equation of the parabola with vertex at the origin and the directrix(M) is y = 2.

We know that the axis is perpendicular to the directrix. Since the directrix is parallel to the x - axis, the axis will parallel to the y - axis.

Since, axis passes through the origin, the equation of the axis is x = 0.

The intersection of the point of the axis and directrix will be (0, 2).

We know that vertex is the midpoint of focus and point on directrix which lies on axis(intersection point).

Let (x₁, y₁) be the focus,

⇒

⇒

⇒ x = 0 and y + 2 = 0

⇒ x = 0 and y = - 2.

The focus is S(0, - 2).

Let P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² + y² + 4y + 4 = y² - 4y + 4

⇒ x² = - 8y

∴The equation of the parabola is x² = - 8y.

Given that we need to find the equation of the parabola whose focus is (5, 2) and having a vertex at (3, 2).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒ m₁ = 0.

Here the axis is parallel to the x-axis, so the directrix should be parallel to the y-axis.

Let us assume the intersection point on directrix is (x₁, y₁).

⇒

⇒

⇒ x + 5 = 6 and y + 2 = 4

⇒ x = 1 and y = 2.

The point on directrix is (1, 2).

We know that the equation parallel to y-axis is x = k. So, the equation of the directrix is x = 1.

Let P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² - 10x + y² - 4y + 29 = x² - 2x + 1

⇒ y² - 8x - 4y + 28 = 0

∴The equation of the parabola is y² - 8x - 4y + 28 = 0.

Given that the cable hangs in the form of a parabola.

It is told that the length of the shortest wire supported is 6.

It is clear that the vertex of the parabola is S(0, 6).

We know that the equation of the parabola having (0, a) as vertex is x² = 4b(y - a)

Let us assume the equation of the parabola is x² = 4b(y - 6).

It is told that the road way is 100m long. We usually give maximum support at the middle of the road i.e. at 50m. At the 50m of the road way, the length of the support wire used is 30m.

It is clear that the point (50, 30) lies on parabola, substituting this point in the equation of the parabola, we get,

⇒ (50)² = 4b(30 - 6)

⇒ 2500 = 4b(24)

⇒ 96b = 2500

The equation of the parabola is .

We need to find the length of the support that is needed to give at the 18m in the roadway.

Let us assume the length of the support is l m,

We have a point (18, l) on the parabola, substituting in the equation we get,

⇒

⇒

⇒ 3.11 = l - 6

⇒ l = 9.11m

∴The length of the support required is 9.11m.

Given that we need to find the equations of the lines joining the vertex to the point whose abscissa is 24 on the parabola y² = 6x.

We know that for a parabola y² = 4ax, the vertex is (0, 0).

So, the vertex of the parabola y² = 6x is (0, 0).

Let us assume the point on parabola be (24, l).

Substituting in the parabola we get,

⇒ l² = 6(24)

⇒ l² = 144

⇒

⇒ l = ±12.

The points on the parabola is (24, ±12).

Let us find the equation of the line passing though the points (0, 0) and (24, 12).

We know that the equation of the straight lines passing through the points (x₁, y₁) and (x₂, y₂) is

⇒

⇒

⇒ x = 2y.

∴The equation of the line is x = 2y.

Let us find the equation of the line passing though the points (0, 0) and (24, - 12).

We know that the equation of the straight lines passing through the points (x₁, y₁) and (x₂, y₂) is

⇒

⇒

⇒ x = - 2y.

∴The equation of the line is x = - 2y.

Given that we need to find the coordinates of points on the parabola y² = 8x whose focal distance is 4.

We know that the focal distance is the distance from the focus to any point on the parabola.

Comparing the given parabola with standard parabola y² = 4ax. We get,

⇒ 4a = 8

⇒ a = 2

⇒ focus = (a, 0) = (2, 0)

We know that point on y² = 4ax is represented by (at², 2at), where t is any real number.

The point on y² = 8x is (2t², 4t)

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒ 16 = 4 + 4t⁴ - 8t² + 16t²

⇒ 16 = 4t⁴ + 8t² + 4

⇒ 4 = t⁴ + 2t² + 1

⇒ 4 = (t² + 1)²

⇒ t² + 1 = 2

⇒ t² = 1

⇒ t = ±1

The points on parabola is,

⇒ (2t², 4t) = (2(±1)², 4(±1))

⇒ (2t², 4t) = (2, ±4)

∴The points on parabola is (2, ±4).

Given that we need to find the length of the line joining the vertex of parabola y² = 4ax and a point on the parabola where the line segment makes an angle θ to the x - axis.

We know that vertex of the parabola is (0, 0).

We know that the equation of the line passing through origin and making angle θ to the x - axis is given by y = (tanθ)x.

Substituting y value in the equation of parabola we get,

⇒ (xtanθ)² = 4ax

⇒ x²tan²θ = 4ax

⇒ xtan²θ = 4a

⇒

⇒ y = tanθx

⇒

⇒

The point on the parabola is .

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ S = 4a²cotθcosecθ.

∴The distance is 4a²cotθcosecθ.

Given that we need to find the equation of the parabola whose focus is (0, 2) and having a vertex at (0, 4).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒ m₁ = ∞.

Here the axis is parallel to the y - axis, so the directrix should be parallel to the x - axis.

Let us assume the intersection point on directrix is (x₁, y₁).

⇒

⇒

⇒ x = 0 and y + 2 = 8

⇒ x = 0 and y = 6.

The point on directrix is (0, 6).

We know that equation parallel to x - axis is y = k. So, the equation of the directrix is y = 6.

Let P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² + y² - 4y + 4 = y² - 12y + 36

⇒ x² + 8y - 32 = 0

∴The equation of the parabola is x² + 8y - 32 = 0.

Given that the line y = mx + 1 is the tangent to the parabola y² = 4x. We need to find the value of m.

Let us substitute the value in the equation of parabola.

⇒ (mx + 1)² = 4x

⇒ m²x² + 2mx + 1 = 4x

⇒ m²x² + (2m - 4)x + 1 = 0

The quadratic will have similar roots if the line is tangent to the parabola.

We know that for a quadratic equation ax² + bx + c = 0 to have equal roots, the condition to be satisfied is b² - 4ac = 0

⇒ (2m - 4)² - 4(m²)(1) = 0

⇒ 4m² - 16m + 16 - 4m² = 0

⇒ 16 - 16m = 0

⇒ 16m = 16

⇒ m = 1

∴The value of m is 1.

Given equation of the parabola is y² = x.

Comparing with the standard form of parabola y² = 4ax,

⇒ The axis of parabola is y = 0.

∴The axis of parabola is y = 0.

Given equation of the parabola is y² + 6y + 2x + 5 = 0

⇒ y² + 6y + 5 = - 2x

⇒ y² + 6y + 9 = - 2x + 4

⇒ (y + 3)² = - 2(x - 2)

Comparing with standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ 4b = 2

⇒

⇒

We know that the distance between the vertex and focus is b.

∴The distance between the vertex and focus is .

Given equation of the parabola is x² - 4x - 8y + 12 = 0

⇒ x² - 4x + 12 = 8y

⇒ x² - 4x + 4 = 8y - 8

⇒ (x - 2)² = 8(y - 1)

Comparing with the standard form of parabola (x - a)² = 4b(y - c) we get,

⇒ 4b = 8

⇒ b = 2

⇒ The equation of the directrix is y - c = - b

⇒ Directrix is y - 1 = - 2

⇒ Directrix is y = - 2 + 1

⇒ Directrix is y = - 1

∴The equation of the directrix is y = - 1.

Given that we need to find the equation of the parabola whose focus is S(0, 0) and directrix(M) is x + y - 4 = 0.

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒

⇒ 2x² + 2y² = x² + y² - 8x - 8y + 2xy + 16

⇒ x² + y² - 2xy + 8x + 8y - 16 = 0

∴The equation of the parabola is x² + y² - 2xy + 8x + 8y - 16 = 0.

Given that we need to find the length of the chord of the parabola y² = 4ax which passes through the vertex and is inclined to axis at . The figure for the parabola is as follows:

We know that the vertex and axis of the parabola y² = 4ax is (0, 0) and y = 0(x - axis) respectively.

We know that the equation of the straight line passing through the origin and inclines to the x - axis at an angle θ is y = tanθx.

⇒

⇒ y = 1.x

⇒ y = x.

The equation of the chord is y = x.

Substituting y = x in the equation of parabola.

⇒ x² = 4ax

⇒ x = 4a.

⇒ y = x = 4a

The chord passes through the points (0, 0) and (4a, 4a).

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒

∴The length of the chord is 4√2a units.

Given that b and c are lengths of the segments of any focal chord of the parabola y² = 4ax.

We know that the length of the latus - rectum is 4a.

We know that the semi - length of the latus - rectum is the harmonic mean of any length of any focal chord.

We know that if a, b, c are in harmonic progression, harmonic mean is given by,

⇒

Here the length of semi-latus rectum is 2a.

⇒

⇒

⇒

Length of the latus - rectum is 4a,

⇒

∴The length of the latus - rectum is .

Given that PSQ is a focal chord of the parabola y² = 8x.

It is also given that SP = 6. We need to find the value of SQ.

Comparing with the standard form of parabola y² = 4ax

⇒ 4a = 8

⇒ a = 2

We know that the semi - length of the latus - rectum is the harmonic mean of any length of any focal chord.

We know that if a, b, c are in harmonic progression, harmonic mean is given by,

⇒

We know that the semi - length of the latus - rectum is 2a = 4.

⇒

⇒

⇒

⇒ SQ = 3

∴The value of SQ is 3.

Given that we need to find the equation of the parabola whose focus is S(- 2, 1) and directrix(M) is x + y - 3 = 0.

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of directrix. We know that the slope of the straight line ax + by + c = 0 is .

⇒

⇒ m₁ = - 1

We know that the product of slopes of the perpendicular lines is - 1.

Let m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒ - 1×m₂ = - 1

⇒ m₂ = 1

We know that the equation of the straight line passing through (x₁, y₁) and having slope m is y - y₁ = m(x - x₁).

⇒ y - 1 = 1(x - (- 2))

⇒ y - 1 = x + 2

⇒ x - y + 3 = 0

On solving the lines x - y + 3 = 0 and x + y - 3 = 0 we get the intersection point to be (0, 3).

Let us assume the vertex be (x₁, y₁).

⇒

⇒ (x₁, y₁) = (- 1, 2)

∴The coordinates of the vertex is (- 1, 2).

Given that we need to find the equation of the directrix of a parabola whose focus is (2, 3) and having a vertex at (- 1, 1).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒ .

We know that the product of slopes of the perpendicular lines is - 1.

Let m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒

⇒

Let us assume the intersection point on directrix is (x₁, y₁).

⇒

⇒

⇒ x + 2 = - 2 and y + 3 = 2

⇒ x = - 4 and y = - 1.

The point on directrix is (- 4, - 1).

We know that equation of the straight line passing through point (x₁, y₁) and slope m is y - y₁ = m(x - x₁).

⇒

⇒ 2(y + 1) = - 3(x + 4)

⇒ 2y + 2 = - 3x - 12

⇒ 3x + 2y + 14 = 0

∴The equation of the directrix is 3x + 2y + 14 = 0.

Given that we need to find the length of the latus - rectum of the parabola y² = 4ax which passes through the point (3, 2).

We know that the length of the latus - rectum of parabola y² = 4ax is 4a.

Substituting the given point in the equation of parabola we get,

⇒ (2)² = 4a(3)

⇒ 4 = 4a(3)

⇒

∴The length of the latus - rectum is .

Given the equation of directrix(M) is x + 5 = 0 and vertex is (- 3, 0).

We know that the directrix and axis are perpendicular to each other. The axis also passes through the vertex.

Let us find the slope of the directrix.

We know that the slope of the line ax + by + c = 0 is .

⇒

⇒ m₁ = ∞.

The slope of the directrix is parallel to the slope of the y - axis. So, the slope of the axis is parallel to the slope of x - axis i.e., 0.

We know that the equation of the line passing through the point (x₁, y₁) and having slope m is (y - y₁) = m(x - x₁)

⇒ y - 0 = 0(x - (- 3))

⇒ y - 0 = 0

⇒ y = 0

On solving the lines x + 5 = 0 and y = 0, we get the intersection point to be (- 5, 0).

We know that vertex is the mid - point of focus and point of intersection of directrix and axis.

Let (x₁, y₁) be the focus.

⇒

⇒

⇒ - 5 + x₁ = - 6 and y₁ = 0

⇒ x₁ = - 1 and y₁ = 0.

The focus is S(- 1, 0).

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ x² + y² + 2x + 1 = x² + 10x + 25

⇒ y² - 8x - 24 = 0

∴The equation of the parabola is y² - 8x - 24 = 0.

Given equation of the parabola is y² - x - 2y + 2 = 0

⇒ y² - 2y + 2 = x

⇒ y² - 2y + 1 = x - 1

⇒ (y - 1)² = (x - 1)

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ 4b = 1

⇒

⇒ The focus is (b + c, a) =

∴The correct option is A

Given that we need to find the vertex of the parabola (y + a)² = 8a(x - a).

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ Vertex of the parabola = (c, a) = (a, - a)

∴The correct option is B

Given the equation of directrix is x + y = 3 and focus is (- 2, 1).

We know that the directrix and axis are perpendicular to each other. The axis also passes through the focus.

Let us find the slope of the directrix.

We know that the slope of the line ax + by + c = 0 is .

⇒

⇒ m₁ = - 1.

We know that the products of the slopes of the perpendicular lines (non - vertical) is - 1. Let us assume the slope of axis be m₂.

⇒ m₁.m₂ = - 1

⇒ - 1.m₂ = - 1

⇒ m₂ = 1.

We know that the equation of the line passing through the point (x₁, y₁) and having slope m is (y - y₁) = m(x - x₁)

⇒ y - 1 = 1(x - (- 2))

⇒ y - 1 = x + 2

⇒ x - y + 3 = 0

On solving the lines x - y + 3 = 0 and x + y = 3, we get the intersection point to be (0, 3).

We know that vertex is the mid - point of focus and point of intersection of axis and directrix.

Let (x₁, y₁) be the vertex of the parabola.

⇒

⇒

⇒ (x₁, y₁) = (- 1, 2)

∴The correct option is C

Given the equation of directrix(M) is x + y = 3a and vertex is (a, 0).

We know that the directrix and axis are perpendicular to each other. The axis also passes through the vertex.

Let us find the slope of the directrix.

We know that the slope of the line ax + by + c = 0 is .

⇒

⇒ m₁ = - 1.

We know that the product of the slopes of the perpendicular lines is - 1.

Let us assume m₂ be the slope of the axis.

⇒ m₁.m₂ = - 1

⇒ (- 1).m₂ = - 1

⇒ m₂ = 1

We know that the equation of the line passing through the point (x₁, y₁) and having slope m is (y - y₁) = m(x - x₁)

⇒

⇒ y = x - a

⇒ x - y - a = 0

On solving the lines x + y = 3a and x - y - a = 0, we get the intersection point to be (2a, a).

We know that vertex is the mid - point of focus and point of intersection of directrix and axis.

Let (x₁, y₁) be the focus.

⇒

⇒

⇒ 2a + x₁ = 2a and a + y₁ = 0

⇒ x₁ = 0 and y₁ = - a.

The focus is S(0, - a).

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ 2x² + 2y² + 4ay + 2a² = x² + y² + 9a² + 2xy - 6ax - 6ay

⇒ x² + y² - 2xy + 6ax + 10ay - 7a² = 0

∴The correct option is B

Given the parametric equations of a parabola x = t² + 1 and y = 2t + 1.

Consider ,

⇒

⇒

⇒

⇒ (y - 1)² = 4(x - 1)

Comparing with the standard form of parabola (y - a)² = 4b(x - c) we get,

⇒ 4b = 4

⇒ b = 1

⇒ The equation of the directrix is x - c = - b

⇒ Directrix is x - 1 = - 1

⇒ Directrix is x = 1 - 1

⇒ Directrix is x = 0

∴The correct option is A

Given that we need to find the equation of the directrix of a parabola whose focus is (2, 3) and having a vertex at (- 1, 1).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒ .

We know that the product of slopes of the perpendicular lines is - 1.

Let m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒

⇒

Let us assume the intersection point on directrix is (x₁, y₁).

⇒

⇒

⇒ x + 2 = - 2 and y + 3 = 2

⇒ x = - 4 and y = - 1.

The point on directrix is (- 4, - 1).

We know that equation of the straight line passing through point (x₁, y₁) and slope m is y - y₁ = m(x - x₁).

⇒

⇒ 2(y + 1) = - 3(x + 4)

⇒ 2y + 2 = - 3x - 12

⇒ 3x + 2y + 14 = 0

∴The correct option is A

Let us assume that the parabola be y² = 4ax.

The parametric equations of the points on the parabola are (at², 2at).

If we assume the points of extremities of the double ordinate of the parabola (at², 2at) and (at², - 2at).

We know that the point of trisection is the point in between these point at a ratio of 2:1 or 1:2.

Let us take the ratio to be 1:2.

Let us assume the point of trisection be (x, y).

⇒

⇒

⇒

Consider y²,

⇒

⇒

⇒

⇒

The locus of point of trisection is parabola.

∴The correct option is C

Given that we need to find the equation of the directrix of a parabola whose focus is (2, 6) and having a vertex at (1, 4).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒ m₁ = 2.

We know that the product of slopes of the perpendicular lines is - 1.

Let m₂ be the slope of the directrix.

⇒ m₁.m₂ = - 1

⇒ 2×m₂ = - 1

⇒

Let us assume the intersection point on directrix is (x₁, y₁).

⇒

⇒

⇒ x₁ + 2 = 2 and y₁ + 6 = 8

⇒ x = 0 and y = 2.

The point on directrix is (0, 2).

We know that equation of the straight line passing through point (x₁, y₁) and slope m is y - y₁ = m(x - x₁).

⇒

⇒ 2(y - 2) = - 1(x)

⇒ 2y - 4 = - x

⇒ x + 2y - 4 = 0

∴The correct option is A

Given equation of the parabola is y² + 6y + 2x + 5 = 0

⇒ y² + 6y + 5 = - 2x

⇒ y² + 6y + 9 = - 2x + 4

⇒ (y + 3)² = - 2(x - 2)

Comparing with standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ 4b = 2

⇒

⇒

We know that the distance between the vertex and the focus is b.

∴The distance between the vertex and the focus is .

∴The correct option is B

Given equation of the parabola is x² - 4x - 8y + 12 = 0

⇒ x² - 4x + 12 = 8y

⇒ x² - 4x + 4 = 8y - 8

⇒ (x - 2)² = 8(y - 1)

Comparing with the standard form of parabola (x - a)² = 4b(y - c) we get,

⇒ 4b = 8

⇒ b = 2

⇒ The equation of the directrix is y - c = - b

⇒ Directrix is y - 1 = - 2

⇒ Directrix is y = - 2 + 1

⇒ Directrix is y = - 1

∴The correct option is C

Given that we need to find the equation of the parabola with focus (0, 0) and directrix x + y = 4

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ 2x² + 2y² = x² + y² + 16 + 2xy - 8x - 8y

⇒ x² + y² - 2xy + 8x + 8y - 16 = 0

∴The correct option is A

Given that the line 2x - y + 4 = 0 cuts the parabola y² = 8x at P, Q. We need to find the midpoint of PQ.

Substituting y = 2x + 4 in the equation of parabola.

⇒ (2x + 4)² = 8x

⇒ 4x² + 16x + 16 = 8x

⇒ 4x² + 8x + 16 = 0

⇒ x² + 2x + 4 = 0

Let x₁, x₂ be the roots. Then, x₁ + x₂ = - 2

Now substituting in the equation of the line we get,

⇒

⇒

⇒ y² - 4y + 16 = 0

Let y₁, y₂ be the roots. Then, y₁ + y₂ = 4

Let us assume P be (x₁, y₁) and Q be (x₂, y₂) and R be the midpoint of PQ.

⇒

⇒

⇒ R = (- 1, 2)

∴The correct option is C

Given that we need to find the length of the chord of the parabola y² = 4ax which passes through the vertex and is inclines to axis at .

We know that the vertex and axis of the parabola y² = 4ax is (0, 0) and y = 0(x - axis).

We know that the equation of the straight line passing through the origin and inclines to the x - axis at an angle θ is y = tanθx.

⇒

⇒ y = 1.x

⇒ y = x.

The equation of the chord is y = x.

Substituting y = x in the equation of parabola.

⇒ x² = 4ax

⇒ x = 4a.

⇒ y = x = 4a

The chord passes through the points (0, 0) and (4a, 4a).

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is .

⇒

⇒

⇒

⇒ l = 4√2 a

∴The correct option is A

Given equation is 16x² + y² + 8xy - 74x - 78y + 212 = 0

We know that for ax² + 2hxy + by² + 2gx + 2fy + c = 0 is a parabola if h² = ab and abc + 2fgh - af² - bg² - ch²≠0

Here a = 16, b = 1, h = 4, g = - 37, f = - 39, c = 212.

⇒ abc + 2fgh - af² - bg² - ch² = (16)(1)(212) + (2)(- 39)(- 37)(4) - (16)(- 39)² - (1)(- 37)² - (212)(4)²

⇒ abc + 2fgh - af² - bg² - ch² = 3392 + 11544 - 24336 - 1369 - 3392

⇒ abc + 2fgh - af² - bg² - ch² = - 14161

⇒ h² = (4)²

⇒ h² = 16

⇒ h² = (16)(1)

⇒ h² = ab

The given curve is parabola.

∴The correct option is B

Given equation of the parabola is y² + 8x - 2y + 17 = 0

⇒ y² - 2y + 17 = - 8x

⇒ y² - 2y + 1 = - 8x - 16

⇒ (y - 1)² = - 8(x + 2)

Comparing with standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ 4b = 8

We know that the that the length of the latus rectum is 4b.

∴The correct option is C

Given equation of the parabola is x² + 8x + 12y + 4 = 0

⇒ x² + 8x + 4 = - 12y

⇒ x² + 8x + 16 = - 12y + 12

⇒ (x + 4)² = - 12(y - 1)

Comparing with standard form of parabola (x - a)² = - 4b(y - c) we get,

⇒ vertex = (a, c) = (- 4, 1)

∴The correct option is A

Given equation of the parabola is (y - 2)² = 16(x - 1)

Comparing with standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒ vertex = (c, a) = (1, 2)

∴The correct option is A

Given equation of the parabola is 4y² + 2x - 20y + 17 = 0

⇒ 4y² - 20y + 17 = - 2x

⇒

⇒

⇒

Comparing with standard form of parabola (y - a)² = - 4b(x - c) we get,

⇒

We know that the that the length of the latus rectum is 4b.

∴The correct option is C

Given equation of the parabola is x² - 4x - 8y + 12 = 0

⇒ x² - 4y + 12 = 8y

⇒ x² - 4x + 4 = 8y - 8

⇒ (x - 2)² = 8(y - 1)

Comparing with standard form of parabola (x - a)² = 4b(y - c) we get,

⇒ 4b = 8

We know that the that the length of the latus rectum is 4b.

∴The correct option is C

Given equation of the parabola is y = 2x² + x

⇒

⇒

⇒

Comparing with standard form of parabola (x - a)² = 4b(y - c) we get,

⇒

⇒

⇒ Focus = (a, b + c) =

∴The correct option is C

Given that the equation of the parabola is x² = 4ay.

We know the parametric equation of the point on the parabola is (2at, at²)

∴The correct option is B

Given that we need to find the equation of the parabola with focus (1, - 1) and directrix x + y + 7 = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on parabola is equidistant from focus and directrix.

We know that the distance between two points (x₁, y₁) and (x₂, y₂) is .

We know that the perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is .

⇒ SP = PM

⇒ SP² = PM²

⇒

⇒

⇒ 2x² + 2y² - 4x + 4y + 4 = x² + y² + 49 + 2xy + 14x + 14y

⇒ x² + y² - 2xy - 18x - 10y - 45 = 0

∴The correct option is D