If two tangents drawn from a point $\left(\alpha ,\beta \right)$ 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 ${\left(10\alpha +5\right)}^2+{\left(16{\beta }^2+50\right)}^2$ equals ______

Let $\alpha =\frac{1}{5}\cos \theta , \beta =\frac{1}{2}\sin \theta$

Equation of tangent to $y^2=4x$ with slope $m$ will be $y=mx+\frac{1}{m}$

Since the tangent passes through $\left(\alpha ,\beta \right)$, so

$\frac{1}{2}\sin \theta =\frac{m}{5}\cos \theta +\frac{1}{m}$

$\Rightarrow m^2\left(\frac{\cos \theta }{5}\right)-m\left(\frac{1}{2}\sin \theta \right)+1=0$

As it is a quadratic equation in $m$ so it will have two roots $m_1$ and $m_2$

Given $m_1=4m_2$

Now, $m_1+m_2=\frac{\frac{1}{2}\sin \theta }{\frac{\cos \theta }{5}}$

$m_1{m}_2=\frac{5}{\cos \theta }$

After eliminating $m_1$ and $m_2$, we get,

$\cos \theta =\frac{-5\pm \sqrt{29}}{2}\Rightarrow {\cos }^2\theta =\frac{54\pm 10\sqrt{29}}{4}$

i.e. $\alpha =\frac{-5\pm \sqrt{29}}{10}\Rightarrow 10\alpha +5=\pm \sqrt{29}$

and ${\beta }^2=\frac{1}{4}{\sin }^2\theta \Rightarrow 16{\beta }^2=-50\pm 10\sqrt{29}$

Hence, ${\left(10\alpha +5\right)}^2+{\left(16{\beta }^2+50\right)}^2=2929$