A) 8
B) 12
C) 16
D) 20
E) 24
Correct Answer: B
Solution :
We have,\[{{a}^{2}}=25\]and\[{{b}^{2}}=16\] \[\therefore \]\[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\] So, the co-ordinates of foci S and S are (3, 0) and\[(-3,0)\]respectively. Let\[P(5\cos \theta ,4\sin \theta )\] be a variable point on the ellipse. Then\[\Delta =area\text{ }of\Delta \text{ }PSS=12\text{ }sin\theta .\] So maximum value of area of\[\Delta PSS\]is 12, since value of\[\sin \theta \]lies between\[-1\]and 1.You need to login to perform this action.
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