Question

If the tangent to the parabola ${{y}^{2}}=x$ at a point $\left( \alpha ,\beta \right),\beta >0$ is also a tangent to the ellipse ${{x}^{2}}+2{{y}^{2}}=1$,\[\] then $\alpha =$
A.$2\sqrt{2}+1$\[\]
B. $\sqrt{2}-1$\[\]
C. $\sqrt{2}+1$\[\]
D. $2\sqrt{2}-1$\[\]

Answer 1

Hint: We differentiate the equation of parabola with respect to $x$ and get the slope at the point $\left( \alpha ,\beta \right)$. Then we put $\left( \alpha ,\beta \right)$ in the equation of parabola and get ${{\beta }^{2}}=\alpha $. We find the intercept of the common tangent putting $\left( \alpha ,\beta \right)$ and the slope in the equation of line $y=mx+c$. We then equate the obtained intercept with intercept in the equation for any tangent to the ellipse which is $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$.\[\]

Complete step by step answer:

We know from differential callus that calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is also the slope of the tangent at that point.
The given equation of a parabola is
\[{{y}^{2}}=x\]

Let us differentiate above equation with respect to $x$ to get the slope at any point
\[\begin{align}
  & \dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( x \right) \\
 & \Rightarrow 2y\dfrac{dy}{dx}=1 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2y} \\
\end{align}\]
So the value of the slope at the given point $\left( \alpha ,\beta \right),\beta >0$ is $m=\dfrac{1}{2y}=\dfrac{1}{2\beta }$. We put the point $\left( \alpha ,\beta \right),\beta >0$ common to both tangent and parabola in equation (1) and get ${{\beta }^{2}}=\alpha $. We use these values in the slope intercept form to find the equation of the tangent to the parabola as
\[\begin{align}
  & y=mx+c \\
 & \Rightarrow \beta =\dfrac{1}{2\beta }\times \alpha +c \\
 & \Rightarrow c={{\beta }^{2}}-\dfrac{1}{2\beta }\times {{\beta }^{2}}=\beta -\dfrac{\beta }{2}=\dfrac{\beta }{2} \\
\end{align}\]

The given equation of ellipse is
\[\begin{align}
  & {{x}^{2}}+2{{y}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}+\dfrac{{{y}^{2}}}{\dfrac{1}{2}}=1 \\
\end{align}\]
We compare above equation with general equation of any hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and get ${{a}^{2}}=1,{{b}^{2}}=\dfrac{1}{2}$. The equation of pair of tangents at any point on the ellipse is given by
\[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]
Here we find the intercept to the tangent to the ellipse as by putting ${{a}^{2}}=1,{{b}^{2}}=\dfrac{1}{2}$.
\[d=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}=\pm \sqrt{1\cdot {{\left( \dfrac{1}{2\beta } \right)}^{2}}+\dfrac{1}{2}}=\pm \sqrt{\dfrac{1}{4{{\beta }^{2}}}+\dfrac{1}{2}}\]

We see that the tangent to parabola and tangent to the ellipse is the same tangent then their intercepts will be equal. So we have
\[\begin{align}
  & c=d \\
 & \Rightarrow \dfrac{\beta }{2}=\pm \sqrt{\dfrac{1}{4{{\beta }^{2}}}+\dfrac{1}{2}} \\
 & \Rightarrow \dfrac{{{\beta }^{2}}}{4}=\dfrac{1}{4{{\beta }^{2}}}+\dfrac{1}{2} \\
\end{align}\]
Let us multiply $4{{\beta }^{2}}$ both side and get
\[\begin{align}
  & \Rightarrow {{\beta }^{4}}-2{{\beta }^{2}}=1 \\
 & \Rightarrow {{\beta }^{4}}-2{{\beta }^{2}}+1=2 \\
 & \Rightarrow {{\left( {{\beta }^{2}}-1 \right)}^{2}}=2 \\
 & \Rightarrow {{\beta }^{2}}=\sqrt{2}+1 \\
 & \Rightarrow \alpha =\sqrt{2}+1\left( \because {{\beta }^{2}}=\alpha \right) \\
\end{align}\]
So the correct option is A.

Note:
We note that the value of slope $\dfrac{1}{2\beta }$ exists because we are given the condition $\beta >0$. We also need to take care of the confusion between the general equation tangents of ellipse from hyperbola which is given by $ y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$.