Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $7$
• E. $9$

Answer 1

The correct option is B

$4$

Explanation for the correct option.

Step 1: Find the slopes of the tangents

Given, curves $x^2=9a(9-y)$ and $x^2=a(y+1)$ intersect orthogonally.

The slope of tangent to curve $x^2=9a(9-y)$.

$$\begin{gathered} 2x=-9a\frac{dy}{dx} \\ {m}_1=-\frac{2x}{9a} \end{gathered}$$

The slope of tangent to curve $x^2=a(y+1)$.

$$\begin{gathered} 2x=a\frac{dy}{dx} \\ {m}_2=\frac{2x}{a} \end{gathered}$$

Step 2: Find the value of $a^2$

As the curves are orthogonal so $m_1{m}_2=-1$.

$$\begin{gathered} \Rightarrow -\frac{2x}{9a}\times \frac{2x}{a}=-1 \\ \Rightarrow \frac{4x^2}{9a^2}=1 \\ \Rightarrow \frac{4\left(9a\right)}{9a^2}=1[\because x^2=9a] \\ \Rightarrow a=4 \end{gathered}$$

Hence, option B is correct.