Question

If \(P\) be any point on the ellipse \(16x^2+25y^2 = 400\) with foci \(S\) and \(S^'\) and area of \(△PSS^'\) is 9 square units, then the abscissa of point \(P\) is...........

• A. \(\frac{7\sqrt{5}}{4}\)
• B. \(\frac{4\sqrt{7}}{5}\)
• C. \(\frac{5\sqrt{7}}{4}\)
• D. \(\frac{10}{7}\)

Hint:

Area of triangle PSS-dash is half base 2ae times the height |y|.

Answer 1

The Correct Option is C

Step 1: Approach
Use the focal geometry and the point on the curve together.

Step 2: Data
The equation $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ gives $a=5$, $b=4$, eccentricity $e=\dfrac35$, foci at $(\pm3,0)$.

Step 3: Height from area
Base $=6$, so $\dfrac12\cdot6\cdot h=9$ gives $h=3$. So the ordinate is $\pm3$.

Step 4: Abscissa
Substitute $y^2=9$: $x^2=25\left(1-\dfrac{9}{16}\right)=\dfrac{175}{16}$, so $|x|=\dfrac{5\sqrt7}{4}$.
This is option (C).

Final Answer:
The ordinate of P is 3, so its abscissa is $\dfrac{5\sqrt{7}}{4}$, option (C). \[ \boxed{\frac{5\sqrt{7}}{4}} \]