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

$\begin{array}{1 1} 1 \\ 0 \\ -6 \\ 6 \end{array}$

Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
• If the tangent of the curves touch orthogonally then the product of their slopes is -1.
Step 1
Equation of the given curves are
$ay+x^2=7\: and \: x^3=y$
Consider the equation,
$ay+x^2=7$
differentiating w.r.t $x$ we get,
$a.\large\frac{dy}{dx}+2x=0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{2x}{a}$
Let this be $m_1 \Rightarrow m_1=-\large\frac{2x}{a}$
$\Rightarrow m_1\: at \: (1,1)=-\large\frac{2}{a}$
Consider the equation,
$y=x^3$
differentiating w.r.t $x$
$\large\frac{dy}{dx}=3x^2$
Let this be $m_2 \Rightarrow m_2 \: at \: (1,1)=3$
Step 2
Since the tangents touch orthogonally,
$m_1 \times m_2 = -1$
$-\large\frac{2}{a} \times 3 = -1$
$\Rightarrow a = 6$
The correct option is D