Illustration 1: If x + y = k is normal to y^{2} = 12x, the find the value of k. (2000)
Solution: We know that if y = mx + c is normal to the parabola y^{2 }= 4ax, then the value of c is = –2am – am^{3}
The given parabola is y^{2} = 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 m_{1}, m_{2}, m_{3} to the parabola y^{2} = 4x. If locus of P with m_{1}m_{2} = α is a part of the parabola itself, then find α. (2003)
Solution: We know that the equation of normal to y^{2} = 4ax is y = mx  2am – am^{3}
Hence, in this case, the equation of normal to y^{2 }= 4x is y = mx  2m – m^{3}
Now, if it passes through (h, k) then we have k = mh – 2m  m^{3}
So, m^{3} + m(2h) + k = 0….. (1)
Here, m_{1 }+ m_{2} + m_{3 }= 0
Hence, m_{1}m_{2 }+ m_{2}m_{3} + m_{3} m_{1} = 2h
And m_{1}m_{2}m_{3} = k, where m_{1}m_{2} = α
Hence, m_{3} = k/ α and it must satisfy eq(1)
k^{3}/ α^{3}  k/ α. (2h) + k = 0
Hence, k^{2} = α^{2}h  2α^{2} + α^{3}
So, y^{2} = α^{2}x  2α^{2} + α^{3}
On comparing this equation with the standard equation y^{2} = 4ax, we get
α^{2} = 4 and 2α^{2} + α^{3} = 0
Hence, the value of α is 2.
Illustration 3: At any point P on the parabola y^{2} – 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^{2 }– 2y  4x + 5 = 0.
This can be rewritten as (y1)^{2} = 4(x1)
And its parametric coordinates are x1= t^{2} and y1 = 2t
So, the point P is P (1 + t^{2}, 1+2t)
Hence, the equation of tangent at P is,
t(y1) = x – 1 + t^{2}, 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 midpoint of RP, then
0 = (h + t^{2} + 1)/2 or t^{2} =  (h+1)
And 1 + t  1/t = (k + 2t + 1)/ 2 or t = 2/ (1k)
Hence, from equations (1) and (2)
4/ (1k)^{2} + (h + 1) = 0
or (k1)^{2}(h+1) + 4 = 0
Hence, the locus of point is
(x+1)(y1)^{2 }+ 4 = 0.
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