Question

If P is any point on the ellipse \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \) and S, S' are its foci, then the maximum area (in sq. units) of \(\Delta SPS' = \)

• A. 15
• B. 12
• C. 6
• D. 25

Hint:

For a triangle with a fixed base, its area is maximized when its height is maximized. In the case of \(\Delta SPS'\) for an ellipse, the base SS' is fixed, so the area is maximum when P is at one of the ends of the minor axis.

Answer 1

The Correct Option is B

Step 1: Find the parameters and foci of the ellipse.
The equation is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).
We have \(a^2=25 \implies a=5\) and \(b^2=9 \implies b=3\).
The distance of the foci from the center is \(c\), where \(c^2 = a^2 - b^2\).
\( c^2 = 25 - 9 = 16 \implies c=4 \).
The foci are S and S' located at \( (4,0) \) and \( (-4,0) \).
Step 2: Formulate the area of the triangle SPS'.
The triangle has its base along the x-axis, the segment SS'.
The length of the base is the distance between the foci, which is \( 2c = 2(4) = 8 \).
Let the point P on the ellipse have coordinates \((x_p, y_p)\).
The height of the triangle is the perpendicular distance from P to the base SS' (the x-axis), which is simply \(|y_p|\).
Area(\(\Delta SPS'\)) = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (8) \times |y_p| = 4|y_p| \).
Step 3: Maximize the area.
To maximize the area, we need to maximize the value of \(|y_p|\) for a point on the ellipse.
From the ellipse equation \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \), the maximum value of \(|y|\) occurs when \(x=0\).
When \(x=0\), \( \frac{y^2}{9} = 1 \implies y^2 = 9 \implies y = \pm 3 \).
The maximum value of \(|y_p|\) is 3. This occurs at the ends of the minor axis, (0,3) and (0,-3).
The maximum area is \( 4 \times 3 = 12 \) square units.