JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

If the ellipse

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

meets the line

\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1

on the

x

-axis and the line

\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1

on the

y

-axis, then the eccentricity of the ellipse is

View answer and explanation

Correct option: A

\frac{5}{7}

Option A (

\frac{5}{7}

) is correct because by finding the values of a and b from the given information, the eccentricity can be calculated using the formula

e = \sqrt{1 - \frac{b^2}{a^2}}

More Questions from ELLIPSE

Practice All Chapter MCQs →

\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