JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

{e_1}

is the eccentricity of the ellipse

\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{7} = 1

and

{e_2}

is the eccentricity of the hyperbola

\frac{{{x^2}}}{9} - \;\;\frac{{{y^2}}}{7} = 1

, then

{e_1} + \;{e_2}

is equal to

View answer and explanation

Correct option: C

\frac{{25}}{{12}}

Option C (

\frac{{25}}{{12}}

) is correct because it is the sum of the eccentricities of the ellipse and hyperbola, calculated using the given equations.

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\frac{x^2}{144}+\frac{y^2}{81}=1

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\mathrm{mM}

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\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

S

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\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

r

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P

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Q

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y=x

Q

R

N P

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P

O