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: Understanding the Concept
Write the ellipse in standard form and find the foci. The base of the triangle is the distance \(SS'\) and its height is the \(y\)-coordinate of \(P\).

Step 2: Standard form
\[ 16x^2+25y^2=400\Rightarrow\frac{x^2}{25}+\frac{y^2}{16}=1 \]
So \(a=5\), \(b=4\), and \(c^2=a^2-b^2=9\), giving \(c=3\). The foci are \((\pm3,0)\) and \(SS'=6\).

Step 3: Area
\[ \text{Area}=\tfrac12\times6\times|y|=3|y|=9\Rightarrow|y|=3 \]

Step 4: Find x
\[ \frac{x^2}{25}=1-\frac{9}{16}=\frac{7}{16}\Rightarrow x^2=\frac{175}{16} \]
\[ x=\pm\frac{5\sqrt7}{4} \]

Step 5: Check the options
The positive value \(\dfrac{5\sqrt7}{4}\) is option (C). Option (A) has the 5 and 7 in the wrong places, and (B) comes from using \(x^2/16\).

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