Applications Of Conic Sections

Given: The focus of a parabolic mirror is at a distance of 6 cm from its vertex. And the mirror is 20 cm deep.

Need to find: Diameter of the mirror.

Here O is the vertex and A is the Focus. So, OA = a = 6 cm.

OD is the deep of the mirror = 20 cm

BC is the diameter of the mirror.

Equation of the parabola is, y² = 4ax

⇒ y² = 24x

The mirror is 20 cm deep. That means the x-coordinate of the points B, C and D is 20

Both the points, B and C are on the parabola. Hence, the points satisfies the equation of the parabola.

Therefore, y² = 24 x 20 = 480

⇒ y = 21.9

So, the coordinate of B is (20, 21.9) and the coordinate of C is (20, -21.9).

Therefore, the diameter of the mirror is = (21.9 + 21.9) cm

= 43.8 cm

Given: Parabolic reflector is 5 cm deep & its diameter is 20 cm

Need to find: Distance of its focus from the vertex.

Reflector is 5 cm deep, i.e., OD = 5 cm

Diameter of the mirror is 20 com, i.e., BC = 20 cm

Let, the equation of the parabola is y² = 4ax, where a is the distance of the focus from the vertex.

The x-coordinate of the points B and C is 5.

D is the middle point of BC which is upon the x-axis.

So, we can say that BD = CD = 10 cm.

So, the coordinate of the point B is (5, 10)

Putting the values of the equation,

y² = 4ax

⇒ 100 = 4a x 5

⇒ 20a = 100

⇒ a = 5

That means, the focus is 5 cm far from the vertex.

Given: A rod of length 15 cm moves with its ends always touching the coordinate axes. A point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis

Need to find: Find the equation of the locus of a point P

Here AB is the rod making an angle with the x-axis.

Here AP = 3.

PB = AB – AP = 12 – 3 = 9 cm

Here, PQ is the perpendicular drawn from the x-axis and RP is the perpendicular drawn from y-axis.

Let, the coordinates of the point P is (x, y).

Now, in the triangle BPQ,

cos =

And in the triangle PAR,

sin =

We know, sin² + cos² = 1

⇒

This is the locus of the point P.

Given: A beam is supported at its ends by supports which are 12 m apart. There is a deflection of 3 cm at the center, and the deflected beam is in the shape of a parabola.

Need to find: How far from the center is the deflection 1 cm

Here EF are the ends of the beam and they are 12 m apart.

IJ is the deflection of 3 cm at the center.

We know, that the distance IF = m = 600 cm and the deflection IJ = FH = 3 cm.

So, the coordinate of the point F is (600, 3)

Let, the equation of the parabola is: x² = 4ay

F point is on the parabola. So, putting the coordinates of F in the equation we get,

x² = 4ay

⇒ 3600 = 4a x 3

⇒ a = 300

Here KL denotes the deflection of 1 cm.

So, at the point L the value of y-coordinate is (3 – 1) = 2

So, by the equation,

⇒ x² = 4ay = 4 x 300 x 2 = 2400

⇒ x = 49 cm.

So, the distance of the point of 1 cm deflection from the center is 49 cm.

Given: Top of the towers are 30 m above the roadway and are 200 m apart. Cable is 5 m above the roadway at center.

Need to find: Length of the vertical supporting cable, 30 m from the center.

A and B are the top of the towers. AE and BF are the height of the towers. H is the center of the bridge. HI is the 5 m above from the roadway.

Let, the equation of the parabola be: x² = 4a(y – b)

Here b = 5. So, x² = 4a(y – 5)

Here, AB = 200 m and BF = 30 m.

So, the coordinate of the point B is (100, 30)

The point is on the parabola.

Hence, x² = 4a(y – 5)

⇒ 10000 = 4a (30 – 5)

⇒ 10000 = 4a x 25

⇒ a = 100

Now we need to find, the length of the vertical supporting cable, 30 m from the center.

The x-coordinate of the point, 30 m from the center, is 30.

So, 30 x 30 = 4a (y – 5)

⇒ 900 = 400 (y – 5)

⇒ y – 5 =

⇒ y =

So, the length of the vertical supporting cable is m = 7.25 m