JEE Main 2019 — Ellipse Question with Solution

$\Rightarrow \left(\alpha , \beta \right)=\left({\beta }^2, \beta \right)$

Tangent to a parabola $y^2=4ax$ at a point $\left(x_1, y_1\right)$ is $y{y}_1=2a\left(x+x_1\right)$

Therefore, tangent to the parabola $y^2=x$ at $\left({\beta }^2, \beta \right)$ is$y\cdot \beta =\frac{1}{2}\left(x+{\beta }^2\right)$

$\Rightarrow y=\frac{x}{2\beta }+\frac{\beta }{2} ...\left(i\right)$

Tangent to parabola is also a tangent to ellipse

$x^2+2y^2=1 \Rightarrow x^2+\frac{y^2}{\left({\displaystyle \frac{1}{2}}\right)}=1$

And, a line $y=mx+c$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if $c^2=a^2{m}^2+b^2$

Using the condition $c^2=a^2{m}^2+b^2,$ we get

$\frac{{\beta }^2}{4}=1^2{\left(\frac{1}{2\beta }\right)}^2+\frac{1}{2}$

$\Rightarrow \frac{{\beta }^2}{4}=\frac{1}{4{\beta }^2}+\frac{1}{2}$

$\Rightarrow {\beta }^4-2{\beta }^2-1=0$

Using Sridharacharya's Formula the roots of a quadratic equation $a{x}^2+bx+c=0, a\ne 0$ are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow {\beta }^2=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\times 1\times \left(-1\right)}}{2\times 1}$

$\Rightarrow {\beta }^2=\frac{2\pm \sqrt{8}}{2}$

$\Rightarrow {\beta }^2=1+\sqrt{2}$ or ${\beta }^2=1-\sqrt{2} ($Neglected as ${\beta }^2$ is positive$)$

$\Rightarrow {\beta }^2=\sqrt{2}+1$

$\Rightarrow \alpha =\sqrt{2}+1.$