JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

e

is eccentricity of ellipse

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

and

e'

is eccentricity of

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a < b} \right),

then

View answer and explanation

Correct option: C

\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1

Option C (

\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1

) is correct because it represents the relationship between the eccentricities of an ellipse and its auxiliary ellipse.

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