Explanation:
(D) : Given curve,
\(2 x^{2}+3 y^{2}=5\)
Differentiating w.r.t. \(\mathrm{x}\) we get -
\(4 x+6 y \cdot \frac{d y}{d x}=0 \Rightarrow 6 y \frac{d y}{d x}=-4 x\)
\(\frac{d y}{d x}=\frac{-4 x}{6 y}\)
\(\therefore\) Slope of the tangent at \((1,1)\) is
\(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=\frac{-4(1)}{6(1)} \Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}=\frac{-4}{6}=\frac{-2}{3}\)
Equation of tangent at \((a, b)\) is
\(y-b=\left(\frac{d y}{d x}\right)_{(a, b)}(x-a)\)
Tangent at \((1,1)\),
\((y-1)=\frac{-2}{3}(x-1)\)
\(3 y-3=-2 x+2\)
\(3 y+2 x-5=0\)
\(2 x+3 y-5=0\)
Slope of the normal at \((1,1)\) is \(\frac{-1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,1)}}=\frac{3}{2}\)
Equation of normal at \((a, b)\) is
\(y-b=\frac{-1}{\left(\frac{d y}{d x}\right)_{(a, b)}}(x-a)\)
Equation of the normal at \((1,1)\) is,
\((y-1)=\frac{3}{2}(x-1)\)
\(2 y-2=3 x-3\)
\(3 x-2 y-1=0\)