Question

If the largest 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 $\alpha$ to:

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

Answer 1

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

$T:y\left(\beta \right)=\frac{1}{2}(x+{\beta }^2)$
$2y\beta =x+{\beta }^2$
$y=\left(\frac{1}{2\beta }\right)x+\frac{\beta }{2}$
$m=\frac{1}{2\beta },C=\frac{\beta }{2}$
$\frac{\beta }{2}=\pm \sqrt{\frac{1}{4{\beta }^2}+\frac{1}{2}}\frac{{\beta }^2}{4}=\frac{1}{4{\beta }^2}+\frac{1}{2}\Rightarrow \frac{{\beta }^2}{4}=\frac{1+2{\beta }^2}{4{\beta }^2}$
$\Rightarrow {\beta }^4-2{\beta }^2-1=0\Rightarrow {({\beta }^2-1)}^2=2\Rightarrow {\beta }^2=\sqrt{2}+1$