Question

Product of the length of the perpendicular segment from the foci on any tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is

• A. a²
• B. b²
• C. a²b²
• D. ab

Answer 1

Answer: (B)

b²

Answer 2

Equation of tangent is y = mx + $\sqrt{a^{2}m^{2} + b^{2}}$.

∴ Product of perpendicular from foci

=$\left| \frac{\sqrt{a^{2}m^{2} + b^{2} +}mae}{\sqrt{m^{2} + 1}} \right|\left| \frac{\sqrt{a^{2}m^{2} + b^{2} -}mae}{\sqrt{m^{2} + 1}} \right| = \frac{a^{2}m^{2} + b^{2} - m^{2}a^{2}e^{2}}{m^{2} + 1}$ = b². [Q b² = a² (1 - e²)]