Explanation:
C Given, the ellipse \(x^2+4 y^2=36\)
Differentiating w.r. t. \(\mathrm{x}\), we get -
\(2 x+8 y \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{-2 x}{8 y}=\frac{-x}{4 y}\)
\(\therefore \frac{d y}{d x}=\frac{-h}{4 k}\)
Then the slope of the normal at \((\mathrm{h}, \mathrm{k})\) is
\(\frac{4 \mathrm{k}}{\mathrm{h}}\)
Let, the equation of the normal is
\(y=m x+c\)
\(\therefore \quad y=\frac{4 k}{h} x+c\)
But this normal passes through \((2,0)\)
\(\therefore 0=\frac{4 \mathrm{k} \times 2}{\mathrm{~h}}+\mathrm{c}\)
\(\therefore \mathrm{c}=\frac{-8 \mathrm{k}}{\mathrm{h}}\)
The equation of normal as
\(y=\frac{4 k}{h} x-\frac{8 k}{h} \quad\), But point \((h, k)\) also lies on the normal
\(\therefore \mathrm{k}=\frac{4 \mathrm{k}}{\mathrm{h}} \cdot \mathrm{h}-\frac{8 \mathrm{k}}{\mathrm{h}}\)
\(\mathrm{k}=4 \mathrm{k}-\frac{8 \mathrm{k}}{\mathrm{h}}\)
\(3 \mathrm{k}=\frac{8 \mathrm{k}}{\mathrm{h}} \Rightarrow \mathrm{h}=\frac{8}{3}\)
Since, point ( \(\mathrm{h}, \mathrm{k}\) ) lies on the ellipse then
\(\mathrm{h}^2+4 \mathrm{k}^2=36\)
\(\therefore \quad 4 \mathrm{k}^2=36-\frac{64}{9}=\frac{260}{9}\)
\(\therefore \mathrm{k}^2=\frac{65}{9}\)
\(\therefore \mathrm{k}=\frac{\sqrt{65}}{3}\)
\(\therefore\) Radius of the circle \((\mathrm{r})=\sqrt{(\mathrm{h}-2)^2+(\mathrm{k}-0)^2}\)
\(=\sqrt{\left(\frac{8}{3}-2\right)^2+\frac{65}{9}}\)
\(=\sqrt{\frac{4}{9}+\frac{65}{9}}=\sqrt{\frac{69}{9}}\)
\(\therefore 12 \mathrm{r}^2=12 \times \frac{69}{9}=4 \times 23=92\)
