A) 16
B) 9
C) 4
D) 3
Correct Answer: B
Solution :
Key Idea: The product of perpendiculars from the foci on any tangent to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is equal to \[{{b}^{2}}.\] We know that the product of perpendiculars from two foci \[{{S}_{1}}\]and \[{{S}_{2}}\]of an ellipse \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1.\]on the tangent at any point P on the ellipse is equal to the square of the semi-minor axis. \[({{S}_{1}}{{M}_{1}})({{S}_{2}}{{M}_{2}})=9.\]You need to login to perform this action.
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