Question

If $ \tan {{\theta }_{1}},\tan {{\theta }_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}}, $ then the chord joining two points $ {{\theta }_{1}} $ and $ {{\theta }_{2}} $ on the ellipse $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ will subtend a right angle at

• A. focus
• B. centre
• C. end of the major axis
• D. end of the minor axis

Answer 1

Answer: (B)

Let $ P(a\cos {{\theta }_{1}},b\sin {{\theta }_{1}}) $ and $ Q(a\cos {{\theta }_{2}},b\sin {{\theta }_{2}}) $ be two points on the ellipse. Then, $ {{m}_{1}}= $ slope of $ OP=\frac{b}{a}\tan {{\theta }_{1}} $ and $ {{m}_{2}}= $ slope of $ OQ=\frac{b}{a}\tan {{\theta }_{2}} $ $ \therefore $ $ {{m}_{1}}{{m}_{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\tan {{\theta }_{1}}\tan {{\theta }_{2}} $ $ =\frac{{{b}^{2}}}{{{a}^{2}}}\times \frac{-{{a}^{2}}}{{{b}^{2}}} $ $ \left[ \therefore \tan {{\theta }_{1}}\tan {{\theta }_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}}(given) \right] $ $ =-1 $ $ \therefore $ $ \angle POQ=\frac{\pi }{2} $ Hence, PQ makes a right angle at the centre of the ellipse.