JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

{e_1}

and

{e_2}

are respectively the eccentricities of the ellipse

\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{4} = 1

and the hyperbola

\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1,

then the relation between

{e_1}

and

{e_2}

View answer and explanation

Correct option: C

2e_1^2 + e_2^2 = 3

Option C (

2e_1^2 + e_2^2 = 3

) is correct because it represents the correct relationship between the eccentricities of the given ellipse and hyperbola.

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

m

\mathrm{M}

\mathrm{mM}

P

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

S

S'.

A

PSS',

A

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

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

r

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

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O