The equation of normal to the curve $2 x^{2}+3 y^{2}-5=0$ at $P(1,1)$ is

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Aicharya · Accepted Answer

Given $2 x^{2}+3 y^{2}-5=0$ $4 x+6 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-2 x}{3 y}$ $\operatorname{At}(1,1), \quad \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-2}{3} \Rightarrow$ Slope of normal $=\frac{3}{2}$ Equation of normal is $\begin{aligned} & y-1=\frac{3}{2}(x-1) \Rightarrow 2 y-2=3 x-3 \ 	herefore \quad & 3 x-2 y-1=0 \end{aligned}$

The equation of normal to the curve $2 x^{2}+3 y^{2}-5=0$ at $P(1,1)$ is

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