Product of lengths of perpendiculars from foci on any tangent to ellipse 3x^2 + 5y^2 = 1 , is

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10-2. Parabola, Ellipse, Hyperbola

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The product of the lengths of perpendiculars from the foci on any tangent to the ellipse $3x^2 + 5y^2 = 1$, is

A

$\frac{1}{5}$

B

$\frac{3}{5}$

C

$\frac{5}{3}$

D

$5$

Solution

It is equal to ${b^2}$

ellipse is $\frac{x^{2}}{\left(\frac{1}{3}\right)}+\frac{y^{2}}{\left(\frac{1}{5}\right)}=1 \Rightarrow b^{2}=\frac{1}{5}$

Standard 11

Mathematics

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