ISI2019-MMA-16

Question

If $S$ and $S’$ are the foci of the ellipse $3x^2 + 4y^2=12$ and $P$ is a point on the ellipse, then the perimeter of the triangle $PSS’$ is

• A. $4$
• B. $6$
• C. $8$
• D. dependent on the coordinates of $P$

Answer 1

The ellipse is $$3x^2 + 4y^2 = 12$$

$$\implies \displaystyle{\frac{x^2}{4} + \frac{y^2}{3} = 1}$$

$$\implies \displaystyle{\frac{x^2}{2^2} + \frac{y^2}{(\sqrt3)^2} = 1}$$

Here, semi-major axis $a = 2$ and semi-minor axis $b = \sqrt3$.

One of the foci can be calculated as $c^2 = a^2 - b^2 = 1 \implies c = 1$.

We know that the sum of the distances of the foci from any point on the ellipse is equal to $2a$.

So, $|PS| + |PS'| = 2a$. The line segments $\overline {PS}$, $\overline {PS'} $, $\overline {SS'} $ form a triangle.

So, $|PS| + |PS'| + |SS'| = 2a + 2c = 2*2 + 2*1 = 6$.

Answer is Option $(B)$.