Question

If tangents drawn to the parabola $(x-1{)}^2-8(y+1)$ which are perpendicular to the variable line $y=px-2p^2-p-1$, 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)

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