Question

The ellipse \(4x^2+9y^2 = 1\) intersects positive Y-axis at point A. If S and S' are its foci, then the area (in sq. units) of \(\Delta SAS^'\) is _____

• A. \(\frac{\sqrt{5}}{12}\)
• B. \(\frac{\sqrt{5}}{18}\)
• C. \(2\sqrt{5}\)
• D. \(3\sqrt{5}\)

Hint:

Find a and b from the standard form, then c = sqrt(a^2 - b^2) and area = c times b.

Answer 1

The Correct Option is B

Step 1: Write in standard form:
\(4x^2+9y^2=1\) becomes \(\dfrac{x^2}{1/4}+\dfrac{y^2}{1/9}=1\). So \(a^2=\dfrac14\) and \(b^2=\dfrac19\). Since \(a>b\), the foci lie on the x-axis.

Step 2: Find the foci and A:
\(c^2=a^2-b^2=\dfrac14-\dfrac19=\dfrac{5}{36}\), so \(c=\dfrac{\sqrt5}{6}\). Foci: \(S(\tfrac{\sqrt5}{6},0)\) and \(S'(-\tfrac{\sqrt5}{6},0)\). On the positive Y-axis, \(x=0\), so \(y=\tfrac13\), and \(A(0,\tfrac13)\).

Step 3: Find the area:
Base \(SS'=2c=\dfrac{\sqrt5}{3}\), height \(=\dfrac13\).
\[ \text{Area}=\dfrac12\cdot\dfrac{\sqrt5}{3}\cdot\dfrac13=\dfrac{\sqrt5}{18} \]

Step 4: Why the other options are wrong.
\(\sqrt5/12\) uses the wrong height, and \(2\sqrt5\), \(3\sqrt5\) result from reading \(a^2=4\), \(b^2=9\) wrongly.

Final Answer:
The area is sqrt 5 / 18. \[ \boxed{\text{(B) }\dfrac{\sqrt5}{18}} \]