Question

Let S & S’ be the foci of an ellipse $b^2(x – 2)^2 + a^2(y – 1)^2 = a^2b^2$ whose eccentricity is e and P is a variable point on the ellipse then

Answer 1

The question is incomplete as stated. If the question were "Find the sum of the distances from a variable point P on the ellipse to its foci S and S'", the answer would be: $2 \times \max(a, b)$

Answer 2

The given equation of the ellipse is: $b^2(x – 2)^2 + a^2(y – 1)^2 = a^2b^2$

To bring it into standard form, divide both sides by $a^2b^2$: $\frac{b^2(x – 2)^2}{a^2b^2} + \frac{a^2(y – 1)^2}{a^2b^2} = \frac{a^2b^2}{a^2b^2}$ $\frac{(x – 2)^2}{a^2} + \frac{(y – 1)^2}{b^2} = 1$ This is the standard equation of an ellipse with center $(h, k) = (2, 1)$. The lengths of the semi-axes are $a$ and $b$.

By the definition of an ellipse, for any point $P$ on the ellipse, the sum of the distances from $P$ to the two foci ($S$ and $S'$) is constant and is equal to twice the length of the semi-major axis.

The semi-major axis is the larger of the two semi-axes lengths, $a$ and $b$. Length of semi-major axis = $\max(a, b)$.

Therefore, the sum of the distances from any variable point $P$ on the ellipse to its foci $S$ and $S'$ is: $PS + PS' = 2 \times (\text{length of semi-major axis})$ $PS + PS' = 2 \times \max(a, b)$

The mention of eccentricity $e$ confirms it is a standard ellipse. If $a > b$, then $a$ is the semi-major axis and $e = \sqrt{1 - b^2/a^2}$. If $b > a$, then $b$ is the semi-major axis and $e = \sqrt{1 - a^2/b^2}$. In either case, the sum of distances is $2 \times \max(a, b)$.