Question

If the foci of the ellipse $\frac{x^2}{9}+y^2=1$ subtend right angle at a point $P$. Then, the locus of $P$ is

• A. $x^2+y^2=1$
• B. $x^2+y^2=2$
• C. $x^2+y^2=4$
• D. $x^2+y^2=8$

Answer 1

The correct option is D

$x^2+y^2=8$

Explanation for the correct option:

Step 1: Solve for the foci of the ellipse

Given that equation of the ellipse is $\frac{x^2}{9}+y^2=1$

We know that the general equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$; where $a$and $b$are the lengths of semi-major and semi-minor axis respectively,

Here, $a=3$ and $b=1$

Eccentricity of the ellipse $e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}$

$\begin{array}{cccl}& & =& \sqrt{1-{\left(\frac{1}{3}\right)}^2}\\ & & =& \frac{\sqrt{8}}{3}\\ & e& =& \frac{2\sqrt{2}}{3}\end{array}$

$\therefore$ Two foci are given by $\left(\pm ae,0\right)$$=\left(\pm 3\times \frac{2\sqrt{2}}{3},0\right)=\left(\pm 2\sqrt{2},0\right)$

Step 2: Solve for the locus of point $P$

Let the co-ordinates of point $P$ be $\left(h,k\right)$

Given that foci subtend right angle at $P$$\Rightarrow$product of slopes$=-1$

Let $m_1,m_2$ be the slopes of the lines joining the points $\left(2\sqrt{2},0\right)$ and $\left(-2\sqrt{2},0\right)$ with $P$ respectively

We know that slope of a line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $m=\frac{y_2-y_1}{x_2-x_1}$

$\begin{array}{cccc}\therefore & m_1& =& \frac{k-0}{h-2\sqrt{2}}\\ & & =& \frac{k}{h-2\sqrt{2}}\\ & m_2& =& \frac{k-0}{h+2\sqrt{2}}\\ & & =& \frac{k}{h+2\sqrt{2}}\end{array}$

Now, according to the given condition $m_1\times m_2=-1$

$\begin{array}{crcl}\Rightarrow & \frac{k}{h-2\sqrt{2}}\times \frac{k}{h+2\sqrt{2}}& =& -1\\ \Rightarrow & \frac{k^2}{h^2-8}& =& -1\\ \Rightarrow & k^2& =& -h^2+8\\ \Rightarrow & h^2+k^2& =& 8\end{array}$

To obtain the locus, replace $\left(h,k\right)$with $\left(x,y\right)$$\Rightarrow x^2+y^2=8$

Hence, the correct answer is option (D) i.e. $x^2+y^2=8$.