Conic Sections Question 57
Question: If the normals at $ P(\theta ) $ and $ Q(\pi /2+\theta ) $ to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ Meet the major axis at G and g, respectively. Then $ PG^{2}+Qg^{2}= $
Options:
A) $ b^{2}(1-e^{2})(2-e^{2}) $
B) $ a^{2}(e^{4}-e^{2}+2) $
C) $ a^{2}(1+e^{2})(2+e^{2}) $
D) $ b^{2}(1+e^{2})(2+e^{2}) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Normal at $ P(\theta ) $ is $ \frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^{2}-b^{2} $ ….(i) Normal at $ P( \frac{\pi }{2}+\theta ) $ is $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ Or $ -\frac{ax}{\sin \theta }-\frac{by}{\cos \theta }=a^{2}-b^{2} $ …(ii) Equation (i) and (ii) meet the major axis at $ G( \frac{(a^{2}-b^{2})cos\theta }{a},0 ) $ and $ g( \frac{(a^{2}-b^{2})sin\theta }{a},0 ) $ Now, $ PG^{2}+Qg^{2}={{{ \frac{(a^{2}-b^{2})cos\theta }{a}-a\cos \theta }}^{2}}+{{(0-bsin\theta )}^{2}} $ $ +{{{ \frac{(a^{2}-b^{2})cos\theta }{a}-a\cos \theta }}^{2}}+{{(0-bcos\theta )}^{2}} $ $ =\frac{{{(a^{2}-b^{2})}^{2}}}{a^{2}}+b^{2}+a^{2} $ $ =a^{2}{ \frac{{{(a^{2}-b^{2})}^{2}}}{a^{4}}+\frac{b^{2}}{a^{2}}+1 } $ $ =a^{2}{ {{( 1-\frac{b^{2}}{a^{2}} )}^{2}}+\frac{b^{2}}{a^{2}}+1 } $ $ =a^{2}(e^{4}+2-e^{2}) $
BETA
