Question

The product of length of perpendicular drawn from the two foci to the tangents at any point on the ellipse $25x^2+4y^2=100$ is

• A. $30\sqrt{5}$
• B. $25\sqrt{2}$
• C. $152$
• D. None of these

Answer 1

Answer: (D)

None of these

Let the point on ellipse be $(2cos\theta ,5\sin \theta )$

The tangent equation at that point is $y-5sin\theta =-\frac{5}{2}cot\theta (x-2cos\theta )$

$\Rightarrow 2y+5cot\theta x=10sin\theta +10cos\theta cot\theta$

$\Rightarrow 2sin\theta y+5cos\theta x=10$

The value of eccentricity $e=\sqrt{1-\frac{4}{25}}=\frac{\sqrt{21}}{5}$

The coordinates of focis are $(0,-\sqrt{21})$ and $(0,\sqrt{21})$

The product of perpendicular distances from focis to tangent line is $\frac{10-2\sqrt{21}sin\theta }{\sqrt{4{\sin }^2\theta +25{\cos }^2\theta }}\times \frac{10+2\sqrt{21}sin\theta }{\sqrt{4{\sin }^2\theta +25{\cos }^2\theta }}$

$\Rightarrow \frac{10-2\sqrt{21}sin\theta }{\sqrt{4{\sin }^2\theta +25{\cos }^2\theta }}\times \frac{10+2\sqrt{21}sin\theta }{\sqrt{4{\sin }^2\theta +25{\cos }^2\theta }}=\frac{100-84{\sin }^2\theta }{4{\sin }^2\theta +25{\cos }^2\theta }=4\left(\frac{25-21{\sin }^2\theta }{25-21{\sin }^2\theta }\right)=4$

Therefore the correct option is $D$