Question

If the coordinates of two points $A,B$ are $\left(\sqrt{7},0\right)$ and $\left(-\sqrt{7},0\right)$ respectively and $P$ is any point on the conic, $9x^2+16y^2=144$, then $PA+PB$ is equal to

• A. $6$
• B. $16$
• C. $9$
• D. $8$

Answer 1

The correct option is D

$8$

Find $PA+PB$:

Step 1: Find the eccentricity

Given an equation of conic, $9x^2+16y^2=144$.

$\frac{x^2}{16}+\frac{y^2}{9}=1$

Which is an equation of an ellipse with $a=4,b=3$.

We know that $b^2=a^2\left(1-e^2\right);e$is eccentricity.

$\begin{array}{rcl}\therefore 9& =& 16\left(1-e^2\right)\\ & \Rightarrow & e^2=1-\frac{9}{16}\\ & \Rightarrow & e^2=\frac{7}{16}\\ & \Rightarrow & e=\pm \frac{\sqrt{7}}{4}\end{array}$

Step 2: Find the value of $PA+PB$

We know that focus of an ellipse is $\left(\pm ae,0\right)$.

Here, $ae=\sqrt{7}$

Therefore, the points $A,B$ are nothing but focus.

$\begin{array}{rcl}\therefore PA+PB& =& 2a\\ & \Rightarrow & PA+PB=2\left(4\right)\\ & =& 8\end{array}$

Hence, option (D) is correct.