Question

If the curve $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then a is equal to

• A. $1$
• B. $-6$
• C. $6$
• D. $0$

Answer 1

The correct option is C $6$

$ay+x^2=7$ ..........$\left(1\right)$

$x^3=y$ ........$\left(2\right)$

at $(1,1)$

Differentiating $\left(1\right)$ wrt x

$a\frac{dy}{dx}+2x=0$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{a}$

Let this be $m_1\Rightarrow m_1=\frac{-2x}{a}$

$m_1$ at $(1,1)=\frac{-2}{a}$

Differentiate $\left(2\right)$ wrt x, then

$\frac{dy}{dx}=3x^2$

Let this be $m_2\Rightarrow m_2$ at $(1,1)=3$

Given that tangent touches orthogonally,

$m_1\cdot m_2=-1$

$\frac{-2}{a}\cdot 3=-1$

$a=6$.