JEE Main 2022 — Circle Question with Solution

Line is passing through intersection of $bx+10y-8=0$ and $2x-3y=0$,

So its equation is given by,

$\left(bx+10y-8\right)+\lambda \left(2x-3y\right)=0 ..........\left(1\right)$

And the line is passing through $\left(1,1\right)$ so $\lambda =b+2$

Now putting the value of $\lambda$ in equation $\left(1\right)$ we get, 

$\Rightarrow \left(3b+4\right)x-\left(3b-4\right)y-8=0$

Now line $\left(3b+4\right)x-\left(3b-4\right)y-8=0$ is tangent to circle $17\left(x^2+y^2\right)=16$

So $\frac{8}{\sqrt{{\left(3b+4\right)}^2+{\left(3b-4\right)}^2}}=\frac{4}{\sqrt{17}}$

$\Rightarrow b^2=2$

Now eccentricity of the ellipse $\frac{x^2}{5}+\frac{{\displaystyle y^2}}{b^2}=1$ will be,

$e^2=1-\frac{b^2}{a^2}$

$\Rightarrow e=\sqrt{1-\frac{2}{5}}=\sqrt{\frac{3}{5}}$