TS EAMCET 4 May 2018 Shift 1 Solved Paper

Let the point of intersection of the required tangent be (h,k) .
Then the equation of tangent through the point to the parabola y2=8xis,
(y2−8x)(k2−8h)=[yk−4(x+h)]2
Now, the equation of tangent to the vertex of parabola is x=0.
So,
y2(k2−8h)=(yk−4h)2
y2k2−8hy2=y2k2 +16h2−8hky
y2−ky+2h=0
Therefore, the roots are y1 and y2.
|y1−y2|=

√k2−8h

1

4=√k2−8h
k2−8h=16
k2=8(h+2)
Take the locus of point (h,k)
y2=8(x+2)