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 $(10\alpha +5{)}^2+(16{\beta }^2$ $+50{)}^2$ equals

Answer 1

Since $(\alpha ,\beta )$ lies on the given ellipse, therefore $25{\alpha }^2+4{\beta }^2=1\cdots \left(1\right)$

Tangent to the parabola, $y=mx+\frac{1}{m}$ passes through $(\alpha ,\beta )$.
So, $\alpha m^2-\beta m+1=0$ has roots $m_1$ and $4m_1$.
$\therefore m_1+4m_1=\frac{\beta }{\alpha }$ and $m_1\cdot 4m_1=\frac{1}{\alpha }$
Gives that $4{\beta }^2=25\alpha \cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$25({\alpha }^2+\alpha )=1$
$\Rightarrow {\alpha }^2+\alpha =\frac{1}{25}$

Now, $(10\alpha +5{)}^2+{(16{\beta }^2+50)}^2$
$=25(2\alpha +1{)}^2+2500(2\alpha +1{)}^2$
$=25\times 101(4{\alpha }^2+4\alpha +1)$
$=25\times 101(\frac{4}{25}+1)$
$=29\times 101=2929$