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: Use the determinant formula:
Area $=\dfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ with points $S(c,0)$, $S'(-c,0)$, $A(0,b)$.

Step 2: Substitute:
Area $=\dfrac12\cdot2c\cdot b=cb$ with $c=\sqrt5/6$ and $b=1/3$, which gives $\dfrac{\sqrt5}{18}$.

Step 3: Pick:
Option B.

Final Answer:
Base 2c = sqrt 5 / 3 and height 1/3 give sqrt 5 / 18. \[ \boxed{\text{(B) }\dfrac{\sqrt5}{18}} \]