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Which one of the following is correct? The eccentricity of the conic
$$
\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}=1,(\lambda \geq 0)
$$
Options:
$$
\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}=1,(\lambda \geq 0)
$$
Solution:
2305 Upvotes
Verified Answer
The correct answer is:
decreases with increase in $\lambda$
Equation of the given conic is an equation of ellipse
$\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}(x \geq 0)$
$\Rightarrow \quad A^{2}=a^{2}+\lambda$ and $B^{2}=b^{2}+\lambda$
Eccentricity, $\mathrm{e}=\sqrt{1-\frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}+\lambda}{\mathrm{a}^{2}+\lambda}}$
$=\sqrt{\frac{a^{2}+\lambda-b^{2}-\lambda}{a^{2}+\lambda}}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}+\lambda}}$
$\lambda$ is in the denominator so, when $\lambda$ increases, the eccentricity decreases.
$\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}(x \geq 0)$
$\Rightarrow \quad A^{2}=a^{2}+\lambda$ and $B^{2}=b^{2}+\lambda$
Eccentricity, $\mathrm{e}=\sqrt{1-\frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}+\lambda}{\mathrm{a}^{2}+\lambda}}$
$=\sqrt{\frac{a^{2}+\lambda-b^{2}-\lambda}{a^{2}+\lambda}}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}+\lambda}}$
$\lambda$ is in the denominator so, when $\lambda$ increases, the eccentricity decreases.
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