If two tangents drawn from a point (alpha, be | vedclass

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Question

$\alpha=\frac{1}{5} \cos 	heta, \beta=\frac{1}{2} \sin 	heta$

Equation of tangent to $y ^{2}=4 x$

$y = mx +\frac{1}{ m }$

It passes through

Vedclass · Accepted Answer

$2929$

Vedclass · Answer

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

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

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