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

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(C) The equation of the tangent to the parabola $y^2 = x$ at point $(\alpha, \beta)$ is given by $y\beta = \frac{x + \alpha}{2}$.Since $(\alpha, \beta

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$\sqrt{2} + 1$

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(C) The equation of the tangent to the parabola $y^2 = x$ at point $(\alpha, \beta)$ is given by $y\beta = \frac{x + \alpha}{2}$.Since $(\alpha, \beta)$ lies on the parabola,$\beta^2 = \alpha$,so the equation becomes $y\beta = \frac{x + \beta^2}{2}$,which simplifies to $y = \frac{1}{2\beta}x + \frac{\beta}{2}$.Here,the slope $m = \frac{1}{2\beta}$ and the intercept $c = \frac{\beta}{2}$.This line is also a tangent to the ellipse $x^2 + 2y^2 = 1$,which can be written as $\frac{x^2}{1} + \frac{y^2}{1/2} = 1$.The condition for a line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.Substituting $a^2 = 1$,$b^2 = 1/2$,$m = \frac{1}{2\beta}$,and $c = \frac{\beta}{2}$:$(\frac{\beta}{2})^2 = 1(\frac{1}{2\beta})^2 + \frac{1}{2}$.$\frac{\beta^2}{4} = \frac{1}{4\beta^2} + \frac{1}{2}$.Multiplying by $4\beta^2$: $\beta^4 = 1 + 2\beta^2$.$\beta^4 - 2\beta^2 - 1 = 0$.Using the quadratic formula for $\beta^2$: $\beta^2 = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.Since $\beta^2 > 0$,we have $\beta^2 = 1 + \sqrt{2}$.Since $\alpha = \beta^2$,we get $\alpha = \sqrt{2} + 1$.

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

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