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Solved Examples on Parabola 

Illustration 1: If x + y = k is normal to y2 = 12x, the find the value of k. (2000)

Solution: We know that if y = mx + c is normal to the parabola y2 = 4ax, then the value of c is = –2am – am3

The given parabola is y2 = 12x

                                        = 4.3x

This gives the value of ‘a’ as 3.

Also, according to the given condition, x + y = k

This can be written as y = (-1)x + k, so m = -1

And we have c = k

Hence, c = k = -2(3)(-1)-3(-1)3 = 9

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Illustration 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 the parabola itself, then find α. (2003)

Solution: We know that the equation of normal to y2 = 4ax is y = mx - 2am – am3

Hence, in this case, the equation of normal to y2 = 4x is y = mx - 2m – m3

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

So, m3 + m(2-h) + k = 0                     ….. (1)

Here, m1 + m2 + m3 = 0

Hence, m1m2 + m2m3 + m3 m1 = 2-h

And m1m2m3 = -k, where m1m2 = α

Hence, m3 = -k/ α and it must satisfy eq(1)

-k3/ α3 - k/ α. (2-h) + k = 0

Hence, k2 = α2h - 2α2 + α3

So, y2 = α2x - 2α2 + α3

On comparing this equation with the standard equation y2 = 4ax, we get

α2 = 4 and -2α2 + α3 = 0

Hence, the value of α is 2.

Illustration 3: At any point P on the parabola y2 – 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. (2004)

Solution: The given parabola is y2 – 2y - 4x + 5 = 0.

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

And its parametric coordinates are x-1= t2 and y-1 = 2t

So, the point P is 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 or let Q be the mid-point of RP, then

0 = (h + t2 + 1)/2 or t2 = - (h+1)

And 1 + t - 1/t = (k + 2t + 1)/ 2 or t = 2/ (1-k)

Hence, from equations (1) and (2)

4/ (1-k)2 + (h + 1) = 0

or (k-1)2(h+1) + 4 = 0

Hence, the locus of point is

(x+1)(y-1)2 + 4 = 0.


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