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**Illustration 1: ****If x + y = k is normal to y ^{2} = 12x, the find the value of k. (2000)**

The given parabola is y

= 4.3x

This gives the value of â€˜aâ€™ 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)

**Illustration 2: Normals are drawn from the point P with slopes m _{1}, m_{2}, m_{3} to the parabola y^{2} = 4x. If locus of P with m_{1}m_{2} = Î± is a part of the parabola itself, then find Î±. (2003)**

Hence, in this case, the equation of normal to y

Now, if it passes through (h, k) then we have k = mh â€“ 2m - m

So, m

Here, m

Hence, m

And m

Hence, m

-k

Hence, k

So, y

On comparing this equation with the standard equation y

Î±

Hence, the value of Î± is 2.

**Illustration 3: At any point P on the parabola y ^{2} â€“ 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)**

This can be rewritten as (y-1)

And its parametric coordinates are x-1= t

So, the point P is P (1 + t

Hence, the equation of tangent at P is,

t(y-1) = x â€“ 1 + t

Hence, y = 1 + t â€“ 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 + t

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

Hence, from equations (1) and (2)

4/ (1-k)

or (k-1)

Hence, the locus of point is

(x+1)(y-1)

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