Question

If curves $x^2=9a(9-y)$ and $x^2=a(y+1)$ intersect orthogonally then value of $a$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

For $x^2=9a(9-y);\frac{dy}{dx}=\frac{-2x}{9a}=m_1$
For $x^2=a(y+1);\frac{dy}{dx}=\frac{2x}{a}=m_2$
$m_1{m}_2=-1$ gives $4x^2=9a^2....\left(1\right)$
also by eliminating $y$ between the equation of curves $x^2=9a.....\left(2\right)$
From (1) and (2) we get $a=4$