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

Option: 1

2929


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{25 x^{2}+4 y^{2}=1}
\mathrm{\frac{x^{2}}{\left(\frac{1}{5}\right)^{2}}+\frac{y^{2}}{\left(\frac{1}{2}\right)^{2}}=1}
Let point on ellipse \mathrm{P:\left ( \frac{1}{5}\cos \theta ,\frac{1}{2} \sin \theta \right )}

\mathrm{y=m x+\frac{1}{m} \quad(\text{Tangent of parabola} \; \; y^{2}=4 x ) }

\mathrm{m \frac{\sin \theta }{2}=\frac{n^{2}\cos \theta }{5}+1 }
\mathrm{\left(2 \cos\theta \right)m^{2}-(5 \sin \theta) m+10=0}

\mathrm{m_{1}+4 m_{1}=\frac{5 \sin \theta}{2 \cos \theta } \quad \& \quad 4 m_{1}\, ^{2}=\frac{5}{\cos \theta }}
               \mathrm{ m_{1}=\frac{ \sin \theta}{2 \cos \theta } }

\mathrm{\therefore \; \frac{ 4\sin^{2} \theta}{4 \cos^{2} \theta }= \frac{5}{\cos \theta } }

       \mathrm{\sin^{2} \theta= 5\cos \theta }
       \mathrm{\cos^{2} \theta +5\cos \theta-1= 0 }
       \mathrm{\cos \theta = \frac{-5\pm \sqrt{29}}{2} } 
\mathrm{\alpha = \frac{1}{5}\cos \theta = \frac{-5\pm \sqrt{29}}{10} }
\mathrm{\left ( 10\, \alpha +5 \right )^{2}= 29 }
\mathrm{16\, \beta ^{2}= -50\pm 10\sqrt{29}}
\mathrm{\therefore \left ( 16\, \beta ^{2}+50 \right )^{2}= 2900}
\mathrm{\therefore \left ( 10\, \alpha +5 \right )^{2}+\left ( 16\,\beta ^{2}+50 \right )^{2}= 2929}

Posted by

Irshad Anwar

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