If the tangent to the parabola y^2 = x at a p | vedclass

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Question

Equation of tangent to the parabola ${y^2} = x$

$At\left( {\alpha ,\beta } \right)$ is $T=0$

$y\beta  = \frac{{x + \alpha }}{2}$

$

Vedclass · Accepted Answer

$\sqrt 2  + 1$

Vedclass · Answer

Equation of tangent to the parabola ${y^2} = x$

$At\left( {\alpha ,\beta } ight)$ 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}$

$\left( {m = \frac{1}{{2\beta }},c = \frac{\beta }{2}} ight)$

This is also a tangent to ellipse ${x^2} + 2{y^2} = 1$

$	herefore 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} ight)^2} = 2$

$ \Rightarrow {\beta ^2} - 1 = \sqrt 2 $

$ \Rightarrow {\beta ^2} = \sqrt 2  + 1$

$\alpha  = {\beta ^2} = \sqrt 2  + 1$

If the tangent to the parabola $y^2 = x$ at a point $\left( {\alpha ,\beta } \right)\,,\,\left( {\beta > 0} \right)$ is also a tangent to the ellipse, $x^2 + 2y^2 = 1$, then $a$ is equal to

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