Question

If the curve ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then a is equal to

(a) 1
(b) −6
(c) 6
(d) 0

Answer 1

(c) 6

$$\begin{gathered} \mathrm{Given}: \\ ay+x^2=7...\left(1\right) \\ {x}^3=y...\left(2\right) \\ \text{Point=}\left(1,1\right) \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) w.r.t.}\mathit{\text{x}}\text{, we get} \\ a\frac{dy}{dx}+2x=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{a} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=\frac{-2}{a} \\ \mathrm{Again},\mathrm{on}\mathrm{d}\text{ifferentiating (2) w.r.t.}\mathit{\text{x}}\text{, we get} \\ 3x^2=\frac{dy}{dx} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=3 \\ \mathrm{It}\mathrm{is}\mathrm{given}\text{that the curves are orthogonal at the given point.} \\ \therefore m_1\times m_2=-1 \\ \Rightarrow \frac{-2}{a}\times 3=-1 \\ \Rightarrow a=6 \end{gathered}$$