Question

If the tangent to the parabola $y^2=x$ at a point $(\alpha ,\beta ),$ $(\beta >0)$ is also a tangent to the ellipse, $x^2+2y^2=1$, then $\alpha$ is equal to :

• A. $\sqrt{2}+1$
• B. $\sqrt{2}-1$
• C. $2\sqrt{2}-1$
• D. $2\sqrt{2}+1$

Answer 1

The correct option is A $\sqrt{2}+1$

$(\alpha ,\beta )$ lies on the parabola $y^2=x$.
$\therefore {\beta }^2=\alpha$
Equation of tangent to the parabola $y^2=x$ at $(\alpha ,\beta )$ is
$y=\left(\frac{1}{2\beta }\right)x+\frac{\beta }{2}\cdots \left(1\right)$

Equation of ellipse is $\frac{x^2}{1}+\frac{y^2}{1/2}=1$
$a=1,b=\frac{1}{\sqrt{2}}$
Condition for $\left(1\right)$ to be tangent to the given ellipse, is
${\left(\frac{1}{2\beta }\right)}^2+\frac{1}{2}=\frac{{\beta }^2}{4}$
$\Rightarrow {\beta }^4-2{\beta }^2-1=0$
$\Rightarrow \alpha ={\beta }^2=\sqrt{2}+1$