Conic Sections Question 327
Question: Which one of the following is correct- The eccentricity of the conic $ \frac{x^{2}}{a^{2}+\lambda }+\frac{y^{2}}{b^{2}+\lambda }=1,(\lambda \ge 0) $
Options:
A) Increases with increase in $ \lambda $
B) Decreases with increase in $ \lambda $
C) Does not change with $ \lambda $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Equation of the given conic is an equation of ellipse $ \frac{x^{2}}{a^{2}+\lambda }+\frac{y^{2}}{b^{2}+\lambda }(x\ge 0) $
$ \Rightarrow A^{2}=a^{2}+\lambda $ and $ B^{2}=b^{2}+\lambda $ Eccentricity, $ e=\sqrt{1-\frac{B^{2}}{A^{2}}}=\sqrt{1-\frac{b^{2}+\lambda }{a^{2}+\lambda }} $
$ =\sqrt{\frac{a^{2}+\lambda -b^{2}-\lambda }{a^{2}+\lambda }} $
$ =\sqrt{\frac{a^{2}-b^{2}}{a^{2}+\lambda }} $
$ \lambda $ is in the denominator so, when $ \lambda $ increases, the eccentricity decreases.
BETA
