If the tangent to the parabola
Question:

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

1. (1) $\sqrt{2}-1$

2. (2) $2 \sqrt{2}-1$

3. (3) $2 \sqrt{2}+1$

4. (4) $\sqrt{2}+1$

Correct Option: , 4

Solution:

Let tangent to parabola at point $\left(\frac{1}{4 m^{2}},-\frac{1}{2 m}\right)$ is

$y=m x+\frac{1}{4 m}$

and tangent to ellipse is, $y=m x \pm \sqrt{m^{2}+\frac{1}{2}}$

Now, condition for common tangency,

$\frac{1}{4 m}=\pm \sqrt{m^{2}+\frac{1}{2}} \Rightarrow \frac{1}{16 m^{2}}=m^{2}+\frac{1}{2}$

$\Rightarrow 16 m^{4}+8 m^{2}-1=0 \Rightarrow m^{2}=\frac{-8 \pm \sqrt{64+64}}{2(16)}$

$=\frac{-8 \pm 8 \sqrt{2}}{2(16)}=\frac{\sqrt{2}-1}{4}$

$\alpha=\frac{1}{4 m^{2}}=\frac{1}{4 \frac{\sqrt{2}-1}{4}}=\sqrt{2}+1$