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JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1
If the normal at any point PP of the ellipse x2a2+y2b2=1(a>b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0) meets the major and minor axes at G{G} and G′G^{\prime} respectively and OF{OF} is the perpendicular drawn from centre O{O} to this normal, then PF⋅PGPF \cdot PG is equal to
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Correct option: A
b2{b}^2
The correct answer is A (b2{b}^2). By finding the coordinates of G and using the perpendicular distance formula, we can derive that PF⋅PG=b2PF \cdot PG = b^2.

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