Question

If two tangents drawn from a point $(\alpha ,\beta )$ lying on the ellipse $25x^2+4y^2=1$ to the parabola $y^2=4x$ 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

Answer: 2929

$\alpha =\frac{1}{5}\cos \theta ,\beta =\frac{1}{2}\sin \theta$
Equation of tangent to $y^2=4x$
$y=mx+\frac{1}{m}$
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 $m_1=4m_2$
$m_1+m_2=\frac{\frac{1}{2}\sin \theta }{\frac{\cos \theta }{5}}$
$m_1{m}_2=\frac{5}{\cos \theta }$
After eliminating $m_1$ and $m_2$
$\cos \theta =\frac{-5\pm \sqrt{29}}{2}$
$\alpha =\frac{-5\pm \sqrt{29}}{10}\Rightarrow 10\alpha +5=\pm \sqrt{29}$
${\beta }^2=\frac{1}{4}{\sin }^2\theta \Rightarrow 16{\beta }^2=-50\pm 10\sqrt{29}$
$(10\alpha +5{)}^2+{\left(16{\beta }^2+50\right)}^2=2929$