Ellipse 2 Question 4
4. If the tangent to the parabola $y^{2}=x$ at a point $(\alpha, \beta),(\beta>0)$ is also a tangent to the ellipse, $x^{2}+2 y^{2}=1$, then $\alpha$ is equal to
(2019 Main, 9 April II)
(a) $\sqrt{2}+1$
(b) $\sqrt{2}-1$
(c) $2 \sqrt{2}+1$
(d) $2 \sqrt{2}-1$
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Answer:
Correct Answer: 4. (a)
Solution:
- Since the point $(\alpha, \beta)$ is on the parabola $y^{2}=x$, so
$ \alpha=\beta^{2} $
Now, equation of tangent at point $(\alpha, \beta)$ to the parabola $y^{2}=x$, is $T=0$
$ \Rightarrow \quad y \beta=\frac{1}{2}(x+\alpha) $
$\left[\because\right.$ equation of the tangent to the parabola $y^{2}=4 a x$ at a point $\left(x _1, y _1\right)$ is given by $\left.y y _1=2 a\left(x+x _1\right)\right]$
$ \begin{array}{ll} \Rightarrow & 2 y \beta=x+\beta^{2} \\ \Rightarrow & y=\frac{x}{2 \beta}+\frac{\beta}{2} \end{array} $
Since, line (ii) is also a tangent of the ellipse
$ x^{2}+2 y^{2}=1 $
$[\because$ condition of tangency of line $y=m x+c$ to ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2} m^{2}+b^{2}$,
here
$ m=\frac{1}{2 \beta}, a=1, b=\frac{1}{\sqrt{2}} \text { and } c=\frac{\beta}{2} $
$\Rightarrow \quad \frac{\beta^{2}}{4}=\frac{1}{4 \beta^{2}}+\frac{1}{\sqrt{2}}$
$\Rightarrow \quad \beta^{4}=1+2 \beta^{2}$
$ \Rightarrow \quad \beta^{4}-2 \beta^{2}-1=0 $
$ \Rightarrow \quad \beta^{2}=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2} $
$\Rightarrow \quad \beta^{2}=1+\sqrt{2}$
$\because \quad \alpha=\beta^{2}=1+\sqrt{2}$
