Q.9. Let a focus of the ellipse E: be S(4,0) and its eccentricity be . If the point P(3, ) lies on E and O is the origin, then the area of is equal to :

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Q.9. Let a focus of the ellipse E:  be S(4,0) and its eccentricity be . If the point P(3, ) lies on E and O is the origin, then the area of  is equal to :

FirstInTest · Accepted Answer

Correct answer: C. Sol. Since focus is S(4,0), the major axis is along the x-axis, so $a > b$.
Focus (ae, 0) = (4, 0) $\Rightarrow ae = 4$.
Given eccentricity $e = \frac{4}{5}$.
So, $a(\frac{4}{5}) = 4 \Rightarrow a = 5$.
Now, $b^2 = a^2(1-e^2) = 5^2(1-(\frac{4}{5})^2) = 25(1-\frac{16}{25}) = 25(\frac{9}{25}) = 9$.
So the equation of the ellipse E is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Point P(3, $\alpha$) lies on E, so $\frac{3^2}{25} + \frac{\alpha^2}{9} = 1 \Rightarrow \frac{9}{25} + \frac{\alpha^2}{9} = 1$.
$\frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \alpha^2 = \frac{9 	imes 16}{25} \Rightarrow \alpha = \pm \frac{12}{5}$.
We can take $\alpha = \frac{12}{5}$.
The vertices of the triangle are O(0,0), P(3, $\frac{12}{5}$), and S(4,0).
Area of $	riangle POS = \frac{1}{2} |x_O(y_P-y_S) + x_P(y_S-y_O) + x_S(y_O-y_P)|$
$= \frac{1}{2} |0(\frac{12}{5}-0) + 3(0-0) + 4(0-\frac{12}{5})|$
$= \frac{1}{2} |0 + 0 - \frac{48}{5}| = \frac{1}{2} 	imes \frac{48}{5} = \frac{24}{5}$.

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