# Solved Examples on Parabola

Parabola is an important head under coordinate geometry. It often fetches good number of questions in various competitive examinations like the JEE.

### Example 1:

Three normals are drawn from the point (c, 0) to the curve y2 = x. Show that c must be greater than ½. One normal is always the x-axis. For what value of ‘c’ are the other two normals perpendicular to each other? (IIT JEE 1991)

### Solution:

We know that the normal of a parabola is y = mx – 2am – am3.

Hence, the equation of normal for y2 = x is y = mx – m/2 – m3/4

Since the normal passes through (c, 0), we have

mc – m/2 – m3/4 = 0

This gives m(c – 1/2 – m2/4) = 0

This yields either m = 0 or m2 = 4(c-1/2)

For m = 0, the equation of normal is y = 0.

Also, m2 ≥ 0, so c-1/2 ≥ 0 or c ≥ ½.

At c = ½, we have m = 0.

For the other normals to be perpendicular to each other, we must have m1m2 = -1.

m2/4(1/2 – c) = 0 has m1m2 = -1.

This means (1/2 - c)/1/4 = -1.

This gives c = ¾.

### Example 2:

Normals are drawn from the point P with slopes m1, m2, m3 to the parabola y2 = 4x. If locus of P with m1m2 = α is a part of parabola itself, then find α. (IIT JEE 2003)

### Solution:

The equation of the normal to the parabola y2 = 4ax.is y = mx – 2am – am3.

Hence, the equation of the normal to the parabola y2 = 4x is y = mx – 2m – m3.

Now, if it passes through (h, k), then we have k =  mh – 2m – m3

This can be written as m3 + m(2-h) + k = 0 …. (1)

Here, m1 + m2 + m3 = 0

m1m2 + m2m3 + m3m1 = 2 - h

Also, m1m2m3 = -k, where m1m2 = α

Now, this gives m3 = -k/α and this must satisfy equation (1).

Hence, we have (-k/α)3 + (-k/α)(2-h) + k = 0

Solving this, we get k2 = α2h – 2α2 + α3

And so we have y2 = α2x – 2α2 + α3

On comparing this equation with y2 = 4x we get

α2 = 4 and -2α2 + α3 = 0

This gives α = 2.

### Example 3:

Find the equation of the parabola whose focus is (3, –4) and directrix is the line parallel to 6x – 7y + 9 = 0 and directrix passes through point (3/2,2).

### Solution:

Let (x, y) be any point on the parabola. Then by definition, the distance between (x, y) and the focus (3, – 4) must be equal to the length of perpendicular from (x, y) on directrix. So first we will find the equation of the directrix.

The line parallel to 6x – 7y + 9 = 0 is  6x – 7y + 6 = 0 …… (1)

Since directrix passes through (3/2,2), this point wil satisfy equation (1) and hence 6(3/2) – 7(2) + k = 0

⇒ k = – 9 + 14 = 5.

Equation of directrix is 6x – 7y + 5 = 0

Now by definition of parabola,

√({(x – 3)+ (y + 4)2}) = (6x – 7y + 5)/√(6+ 72)

⇒ 85{(x – 3)2 + (y + 4)2} = (6x – 7y + 5)2

⇒ 49x2 + 36y2 + 84xy – 570x + 750y + 2100 = 0

### Example 4:

Find the equation of the parabola whose directrix makes an isosceles right angled triangle of area 4 square units with the axis in the 3rd quadrant and focus is on the line y = x, 2 units away from the origin.

### Solution:

First we find the equation of directrix. Let the directrix form the isosceles triangle OAB with OA = OB = a.

Then according to the given condition, ar (? OAB) = 4

⇒ 1/2 a2= 4

⇒ a = ± 2√2 {? the triangle in 3rd quadrant so =-2√2}

Therefore the co-ordinate, of A and B are (–2√2, 0) and (0, –2√2) respectively.

So, equation of directrix

(y – 0) = ((0+2√2)/(-2√2-0)) (x + 2√2) ⇒ x + y + 2√2 = 0

Now the focus S is on line y = x and 2 units away from the origin i.e. OS = 2 ⇒ point (√2, √2) by definition of parabola, we have

√((x-√2)+ (y-√2)2) = |(x+y+2√2)/√(12+12)|

⇒ x2 + y2– 8√2x – 8√2y – 2xy = 0.

### Example 6:

At any point P on the parabola y- 2y – 4x + 5 = 0, a tangent is drawn which meets the directrix at Q. Find the locus of point R, which divides QP externally in the ratio 1/2:1.(IIT JEE 2004)

### Solution:

The given parabola is y- 2y – 4x + 5 = 0.

This can be rewritten as (y-1)2 = 4(x-1)

Its parametric coordinates are x-1 = t2 and y -1 = 2t and hence we have P(1 + t2, 1 + 2t)

Hence, the equation of tangent at P is

t(y-1) = x – 1 + t2, which meets the directrix x = 0 at Q.

Hence, y = 1 + t – 1/t or Q(0, 1 + t – 1/t)

Let R(h, k) be the point which divides QP externally in the ratio 1/2:1. Q is the mid-point of RP

So, we have 0 = (h + t+ 1)/2 which gives t2 = – (h + 1) …. (1)

and 1 + t – 1/t = (k + 2t + 1)/2 which gives t = 2/(1- k) ….. (2)

Hence, from equations (1) and (2) we get, 4/(1-k)2 + (h + 1) = 0\or (k-1)2(h +1) + 4 = 0

Hence, the locus of a point is

(x + 1)(y - 1)2 = 0.

### Example 7:

Find the equation of the common tangents to the parabola y2 = 32x and x2 = 108y.

### Solution:

The equation of the tangent to the parabola y2 = 4ax, is

y = mx + a/m …… (1)

The equation of tangent to the parabola y2 = 32x …… (2) is

y = mx + 8/m …… (3)

If this line given by (3) is also a tangent to the parabola x2 = 108y, then (3) meets x2 = 108y …… (4)  in two coincident points

⇒ Substituting the value of y from (3) in (4) we get  x2 = 108 [mx + (8/m)]

⇒ mx2 – 108m2x – 864 = 0

The roots of this quadratic are equal provided b2 = 4ac.

i.e. (– 108 m2)2 = 4m (–864)

⇒ m = (-2)/3 (m ≠ 0, from geometry of curves)

Substituting their value of m in (3), the required equation is

y = (-2)/3 x + 8/(–2/3)

y = (-2)/3 x – 12

⇒ 2x + 3y + 36 = 0

### Example 8:

Find the angle of intersection of the parabola y2 = 8x and x2 = 27y.

### Solution:

The given parabolas are y2 = 8x …… (1)

and x2 = 27y …… (2)

Solving (1) and (2) we get

(x2/27)2 = 8x

⇒ x4 = 5832 x

⇒ x4 – 5832 x = 0

⇒ x(x3 – 5832) = 0

⇒ x = 0, x = 18

Substituting these values of x in (2) we get y = 0, 12

The point of intersection are (0, 0), (18, 12).

Now, we consider these points one by one

At the point (0, 0)

y2 = 8x ⇒ dy/dx = 4/y ⇒ dy/dx|(0,0) = ∞

x= 27y ⇒ dy/dx = 2x/27 ⇒ dy/dx|(0,0) = 0

⇒ The two curves intersect at the point (0, 0) at right angle.

At the point (18, 12)

y= 8x ⇒ dy/dx|(18,12) = 4/12 = 1/3 = m1 (say)

x= 27y ⇒ dy/dx = 2x/27 ⇒ dy/dx|(18,12) = (2×18)/27 = 4/3 = m2 (say)

Let θ be the angle at which the two curves intersect at the point (18, 12)

Then tan θacute = |(m- m1)/(1 + m1 m2)| = 3/13

θacute = tan-1 (3/13).

### Example 9:

Prove that (x + a)2 = (y- 4ax), is the locus of the point of intersection of the tangents to the parabola y= 4ax, which includes an angle π/4.

### Solution:

Let two tangent to the parabola y2 = 4ax …… (1)

be yt1 = x + at12…… (2)

and yt2 = x + at2 …… (3)

Let the point of intersection of the tangent be (x1, y1) then solving equation (1) and (2) we get, x1 = at1t2

y1 = a(t1 + t2)

Also the slope of these tangents are 1/t1 and 1/t2

∴ If α be the angle between these two tangents then

tan α = + (m- m2)/(1 + m1m2) = (±(1/t1 -1/t2))/(1 + 1/t1 × 1/t2).

⇒ tan α = ± ((t- t1)/(1 + t2t1))

We are given α = π/4

∴ tan π/4 = 1 = ± ((t- t1)/(1 + t2t1))

⇒ (1 + t1t2)2 = (t2 – t1)

⇒ {1 + (x1/a)2} = (t1 – t2)2 – 4t1t2 = (y1/a)- 4x1/a

⇒ (x1 + a)2 = y12– 4ax1

⇒ Required locus of (x1, y1) is (x + a)2 = (y- 4ax).

### Example 10:

If two tangents to a parabola intercept a constant length on any fixed tangent, find the locus of their point of intersection.

### Solution:

Let yt = x + at2 …… (1)

be a fixed tangent to the parabola y= 4ax …… (2)

Let the other two tangent to (2) be yt1 = x + at12…… (3)

And yt2 = x + at22…… (4)

The point of intersection of (1) with (3) and (4) are P {att1, a(t1 + t2)} and Q{at t2, a (t + t2)}

Given PQ is constant or PQ2= constant

⇒ a2t2 (t1 – t2)+ a2(t1 – t2)2 = constant

⇒ a2(t2 + 1) (t1 – t2)2= constant

⇒ (t1 – t2)2 = constant since t is constant as (1) is a fixed tangent.

⇒ (t1 – t2)2 = c (say) …… (5)

Let (x1, y1) be the point of intersection of (3) and (4),

then x1 = at1t2 and y1 = a(t1 + t2) …… (6)

We know that

(t2 – t1)2 = (t1 + t2)2

From equation (5), (6) and 97) we get

c = (y1/a)2- 4x1/a  or y1= 4x1 a – a2

c = 4a (x+ 1/4 ac)

⇒ the locus of (x1, y1) is  y= 4a (x + ¼ ac)

⇒ The required locus is a parabola whose latus rectum is 4a i.e. equal to latus rectum of y2 = 4ax.

### Illustration 11:

Find the locus of the poles of normals to parabola y= 4ax.

### Solution:

Any normal to the parabola y= 4ax is …… (1)

y = mx – 2am – am3 …… (2)

Let (x1, y1) be the pole of (2) with respect to (1), then (2) is the polar of (x1, y1) w.r.t (1)  i.e.

yy1 = 2a (x + x1) comparing (2) & (3), we get  2a/m = y1/1 = (2ax1)/(-2am – am3)

Hence we get  x1 = –2a – am2…… (4)

and y1 = 2a/m …… (5)

Eliminating m between (4) & (5) we get y12(x1 + 2a) + 4a3 = 0

∴ The required locus of (x1, y1) is  (x + 2a)y+ 4a3 = 0

### Illustration 12:

If the line x -1 = 0 is the directrix of the parabola y2 – kx + 8 = 0, then one of the values of k is (IIT JEE 2000)

### Solution:

The given parabola is y2 – kx + 8 = 0.

This can be written as y2 = kx - 8

This gives y2 = k (x - 8/k)

Shifting the origin we get Y2 = kX, where Y = y and X = x-8/k

The directrix of the standard parabola is X = -k/4

The directrix of the original parabola is x = 8/k – k/4

Also, x = 1 coincides with x = 8/k – k/4.

On solving these we get k = 4.

### Illustration 13:

A tangent to the parabola y2 + 12x = 0 cuts the parabola y2 = 4ax at P and Q. Find the locus of middle points of PQ.

### Solution:

Any tangent to the parabola y2 = – 4bx is y = mx – b/m

y = mx – 3/m

Let (x1, y1) be the mid point of PQ, where P and Q are point of intersection o line (1) and y2 = 4ax

Equation of chord PQ is

S1 = T

y12 – 4ax1 = y1y – 2a(x + x1)

y1y – 2ax – y12 + 2 ax1 = 0 …… (2)

Equation (1) can be written as

my – m2x + 3 = 0 …… (3)

equation (2) and (3) represent the same line

⇒ m/y1 = m2/2a = 3/(2ax- y12) …… (4)

⇒ m = 2a/y1 (from 4)

Again, from (4), we get

m/y1 = 3/(2ax- y12 )

⇒ 2a/y1 = 3/(2ax- y12 )

⇒ Locus of (x1, y1) is  4a2x = y2 (3 + 2a)

### Illustration 14:

The point of intersection of the tangents at the ends of the latus rectum of the parabola y2 = 4x is ..?(IIT JEE 1994)

### Solution:

The coordinates of the xetremities of the latus rectum of y2 = 4ax are (1, 2) and (1, -2).

The equations of tangents at these points are gievn by

y.2 = 4(x+1)/2

This gives 2y = 2(x + 1)...... (1)

and y(-2) = 4(x+1)/2

which gievs -2y = 2(x + 1) ….. (2)

The points of intersection of these tangents can be obtained by solving these two equatiosn simultaneously.

Therefore, -2(x +1) = 2(x + 1)

which gives 0 = 4(x +1)

this yields x = -1 and y = 0.

Hence, the required point is (-1, 0).

### Illustration 15:

The normal at any point P of the y2 = 4ax meets the axis in G and to the tangent at the vertex at H. If A be the vertex and the rectangle AGQH be completed, prove that the locus of Q is x= 2ax2 + ay2.

### Solution:

Any normal to the parabola y= 4ax … (1)

is y = mx – 2am – am3 … (2)

This normal (2) meets the axis y = 0 of (1) in G and the tangent at the vertex i.e. x = 0 in H

∴ The coordinates of G and H are (2a + am2, 0) and (0, –2am – am3) respectively. Also the vertex A is (0, 0). Let Q be (x1, y1)

Given that AGQH is a rectangle. AQ and GH are its diagonals and therefore there mid points are same. Now the mid point of AQ is (1/2 x1,1/2 y1 ) and that of GH is  [1/2 (2a + am2+0) 1/2 (0-2am-am3 ) ]

i.e. [1/2 (2a+am2 )-1/2 (2am+am3 ) ]

⇒ The mid points coincide so we have

1/2 x1 = 1/2 (2a + am2), 1/2 y1 = -1/2 (2am + am3)

or x1 = 2a + am2, y1 = – (2am + am3)

The required locus of Q is obtained by eliminating m between these.

Now y12 = (m2a2) (2 + m2)2 = a (m2a) [2 + am2/a]2 = a(x1 – 2a) (x1/a)2

⇒ ay12 = (x1 – 2a) x12

So the locus of Q is ⇒ ay2 + 2ax2 = x3

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