Question

If the curves $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$ then $a=?$

• A. $1$
• B. $-6$
• C. $6$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $6$

Concept : If the tangent of the curves

touch orthogonally then the product

of their slopes = -1

$ay+x^2=7$

Differentiating w.r.t. x,

${}^a\left(\frac{dy}{dx}\right)+2x=0$

$\therefore \left(\frac{dy}{dx}\right)=\frac{-2x}{a}$

$m_1=(\frac{dy}{dx}{)}_{1,1}=\frac{-2}{a}.$ at (1,1)

$y=x^3$ __ (2)

Differentiating w.r.t.x

$\frac{dy}{dx}=3x^2$ slope $\left(m_2\right)=(\frac{dy}{dx}{)}_{(1,1)}=3$

since, tangents touch orthogonally,

$m_1.m_2=-1$

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

$\therefore a=6$