Q. 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 a is equal to :
Solution:
$T:y \left(\beta\right) = \frac{1}{2} \left(x+\beta^{2}\right)$
$ 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} $
$ \frac{\beta^{2}}{4} = \frac{1}{4\beta^{2}} + \frac{1}{2} \frac{\beta^{2}}{4} = \frac{1+2\beta^{2}}{4\beta^{2}}$
$ \Rightarrow \beta^{4} -2\beta^{2} -1 = 0 $
$ \left(\beta^{2} -1\right)^{2} = 2$
$ \beta^{2} -1 = \sqrt{2} $
$ \beta^{2} = \sqrt{2}+1 $
