Question

If the tangent at any point $ P $ on the ellipse $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ meets the tangents at the vertices $ A $ and $ A' $ in $ L $ and $ L' $ respectively, then $ AL\cdot \text{ }A'L' $ is equal to

• A. $ a+b $
• B. $ {{a}^{2}}+{{b}^{2}} $
• C. $ {{a}^{2}} $
• D. $ {{b}^{2}} $

Answer 1

Answer: (D)

Let $ P(a\cos \theta ,\,\,b\sin \theta ) $ be any point on the ellipse. Then, equation of the tangent at $ P $ is $ \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 $
It cuts the lines $ x=a $ and $ x=-a $ $ L\left( a,\,\,\frac{b(1-\cos \theta )}{\sin \theta } \right) $ and $ L'\left( -a,\,\,\frac{b(1+\cos \theta )}{\sin \theta } \right) $ respectively.
Since, $ A $ and $ A' $ are the vertices of given ellipse.
Therefore, coordinates $ A(a,\,\,0) $ and $ B(0,\,\,-a) $ .
$ \therefore $ $ AL=\frac{b(1-\cos \theta )}{\sin \theta } $ and $ AL'=\frac{b(1+\cos \theta )}{\sin \theta } $