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JEE Main Mock Test Series 2020 & Previous Year Papers

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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 y= 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

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 y= 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, m+ m2 + m= 0
Hence, m1m+ 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 y– 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)+ 4 = 0.

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