Question

If tangents drawn to the parabola $(x-1)^{2}-8(y+1)$ which are perpendicular to the variable line $y=p x-2 p^{2}-p-1$, where $p$ is a parameter, then point of intersection of these tangents to the variable line lie on the curve, which is

• A. $y+3=0$
• B. $y=3$
• C. $x=3$
• D. $x+3=0$

Answer 1

Answer: (A)

$ ( x -1)^{2}=8( y +1) \Rightarrow X ^{2}=8 Y$
$y = px -2 p ^{2}- p -1$
$y +1= p ( x -1)-2 p ^{2} $
$Y = pX -2 p ^{2} $
$\Rightarrow$ which is tangent to the parabola $X ^{2}=8 Y$
Tangents is perpendicular to the given variable line which are tangents to the parabola, meet on the directrix of the parabola.
$\therefore$ Directrix is, $y +1+2=0 \Rightarrow y +3=0$