Question

Let the line $y=mx$ and the ellipse $2x^2+y^2=1$ intersect a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at $\left(-\frac{1}{3\sqrt{2}},0\right)$ and $\left(0,\beta \right)$, then $\beta$ is equal to

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{2}{3}$
• C. $\frac{2\sqrt{2}}{3}$
• D. $\frac{\sqrt{2}}{3}$

Answer 1

The correct option is D

$\frac{\sqrt{2}}{3}$

Explanation for the correct option:

Step-1: Normal of the ellipse:

Given the line $y=mx$ and ellipse $2x^2+y^2=1$.

Let $P$ be the point $\left(x_1,y_1\right)$.

Normal equation on ellipse $\frac{\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{-}\mathbf{}\frac{{\mathbf{b}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{.}$

The equation of normal at $P$ is defined as, $\frac{x}{2x_1}-\frac{y}{y_1}=-\frac{1}{2}$.

It is given that the line passes through the point $\left(-\frac{1}{3\sqrt{2}},0\right)$.

Now substitute $x$ as $-\frac{1}{3\sqrt{2}}$ and $y$ as $0$ in the equation of normal.

$\begin{array}{rcl}& \Rightarrow & \frac{-{\displaystyle \frac{1}{3\sqrt{2}}}}{2x_1}-\frac{0}{y_1}=-\frac{1}{2}\\ & & \end{array}$

$\Rightarrow$ $\begin{array}{rcl}-\frac{1}{6\sqrt{2}{x}_1}& =& -\frac{1}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{3\sqrt{2}{x}_1}& =& 1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{x}_1& =& \frac{1}{3\sqrt{2}}\end{array}$

Step-2 finding the value of $y_1$:

Consider the equation of an ellipse $2x^2+y^2=1$at $\left(x_1,y_1\right)$

$\Rightarrow 2{x_1}^2+{y_1}^2=1$

substitute $x_1$ as $\frac{1}{3\sqrt{2}}$ in the above equation.

$\begin{array}{rcl}& \Rightarrow & 2{\left(\frac{1}{3\sqrt{2}}\right)}^2+y_1^2=1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2\left(\frac{1}{9\times 2}\right)+y_1^2& =& 1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{y}_1& =& \frac{2\sqrt{2}}{3}\end{array}$

Squaring both sides.

$\Rightarrow$. $\begin{array}{rcl}{y}_1^2& =& \frac{8}{9}\end{array}$

Since $P$ lies on the normal of the ellipse $\beta =\frac{y_1}{2}$.

Substitute $y_1$ as $\frac{2\sqrt{2}}{3}$ in the $\beta =\frac{y_1}{2}$.

$$\begin{gathered} \beta =\frac{{\displaystyle \frac{2\sqrt{2}}{3}}}{2} \\ \Rightarrow \beta =\frac{2\sqrt{2}}{6} \\ \Rightarrow \beta =\frac{\sqrt{2}}{3} \end{gathered}$$

Therefore, $\beta$ is equal to $\frac{\sqrt{2}}{3}$.

Hence, option D is the correct answer.