Question

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25 x^{2}+4 y^{2}=1$ to the parabola $y^{2}=4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2} $ equals ___

Answer 1

$\alpha=\frac{1}{5} \cos \theta, \beta=\frac{1}{2} \sin \theta$
Equation of tangent to $y ^{2}=4 x$
$y = mx +\frac{1}{ m }$
It passes through $(\alpha, \beta)$
$\frac{1}{2} \sin \theta= m \frac{1}{5} \cos \theta+\frac{1}{ m } $
$ m ^{2}\left(\frac{\cos \theta}{5}\right)- m \left(\frac{1}{2} \sin \theta\right)+1=0$
It has two roots $m_{1}$ and $m_{2}$ where $m_{1}=4 m_{2}$
$ m _{1}+ m _{2}=\frac{\frac{1}{2} \sin \theta}{\frac{\cos \theta}{5}}$
$ m _{1} m _{2}=\frac{5}{\cos \theta}$
After eliminating $m_{1}$ and $m_{2}$
$\cos \theta=\frac{-5 \pm \sqrt{29}}{2}$
$\alpha=\frac{-5 \pm \sqrt{29}}{10} \Rightarrow 10 \alpha+5=\pm \sqrt{29}$
$\beta^{2}=\frac{1}{4} \sin ^{2} \theta \Rightarrow 16 \beta^{2}=-50 \pm 10 \sqrt{29}$
$(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}=2929$