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यदि परवलय $y ^{2}= x$ के एक बिन्दु $(\alpha, \beta),(\beta>0)$ पर, स्पर्श रेखा, दीर्घवृत्त $x ^{2}+2 y ^{2}=1$ की भी स्पर्श रेखा है, तो $\alpha$ बराबर है
$2\sqrt 2 + 1$
$\sqrt 2 - 1$
$\sqrt 2 + 1$
$2\sqrt 2 - 1$
Solution
Equation of tangent to the parabola ${y^2} = x$
$At\left( {\alpha ,\beta } \right)$ is $T=0$
$y\beta = \frac{{x + \alpha }}{2}$
$ \Rightarrow y\beta = \frac{{x + {\beta ^2}}}{2}\,$ ($\because$ ${\beta ^2} = \alpha $)
$ \Rightarrow y = \frac{1}{{2\beta }}x + \frac{\beta }{2}$
$\left( {m = \frac{1}{{2\beta }},c = \frac{\beta }{2}} \right)$
This is also a tangent to ellipse ${x^2} + 2{y^2} = 1$
$\therefore C = \pm \sqrt {{a^2}{m^2} + {b^2}} $
$ \Rightarrow \frac{\beta }{2} = \pm \sqrt {\frac{1}{{4\beta }} + \frac{1}{2}} $
$ \Rightarrow \frac{{{\beta ^2}}}{4} = \frac{1}{{4\beta }} + \frac{1}{2}$
$ \Rightarrow {\beta ^4} – 2{\beta ^2} – 1 = 0$
$ \Rightarrow {\left( {{\beta ^2} – 1} \right)^2} = 2$
$ \Rightarrow {\beta ^2} – 1 = \sqrt 2 $
$ \Rightarrow {\beta ^2} = \sqrt 2 + 1$
$\alpha = {\beta ^2} = \sqrt 2 + 1$
