JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

Let

e_1

and

e_2

be the eccentricities of the ellipse

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

and hyperbola

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

respectively. If the point

\left(e_1, e_2\right)

lies on the ellipse

15 x^2+3 y^2=k

, then the value of

k

View answer and explanation

Correct option: C

Option C (16) is correct because it is the value obtained by substituting the eccentricities of the ellipse and hyperbola into the given equation.

More Questions from ELLIPSE

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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