Mathematics - Ellipse Question with Solution | TestHub

We have equation of circle $x^2+y^2=\frac{1}{2}$ and equation of parabola $y^2=4x$\

Let the equation of common tangent of parabola and circle is $y=mx+\frac{1}{m}$

Since radius of circle $=\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}=\left|\frac{0+0+\frac{1}{m}}{\sqrt{1+m^2}}\right|$

$\Rightarrow {\mathrm{m}}^4+{\mathrm{m}}^2-2=0\Rightarrow \mathrm{m}=\pm 1$

∴ Equation of common tangents are $\mathrm{y}=\mathrm{x}+1$ and $\mathrm{y}=-\mathrm{x}-1$.

Intersection point of common tangent at $\mathrm{Q}(-1,0)$.

∴ Equation of ellipse $\frac{{\mathrm{x}}^2}{1}+\frac{{\mathrm{y}}^2}{1/2}=1$

Where, ${\mathrm{a}}^2=1,{\mathrm{b}}^2=1/2$

Now, eccentricity $(e)=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$

and length of latus rectum $=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2\left(\frac{1}{2}\right)}{1}=1$

∴ Area of shaded region $=2{\int }_{1/\sqrt{2}}^1\frac{1}{\sqrt{2}}\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}$

$$\begin{array}{rlrl}& =\sqrt{2}{\left[\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}+\frac{1}{2}{\sin }^{-1}\mathrm{x}\right]}_{1/\sqrt{2}}^1& =\sqrt{2}\left[\left(0+\frac{\pi }{4}\right)-\left(\frac{1}{4}+\frac{\pi }{8}\right)\right]& \end{array}$$

$$=\sqrt{2}\left(\frac{\pi }{8}-\frac{1}{4}\right)=\frac{\pi -2}{4\sqrt{2}}$$