JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

Let the equations of two ellipses be

E_{1}: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1 \text { and } E_{2}: \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1,

If the product of their eccentricities is

\frac{1}{2}

, then the length of theminor axis of ellipse

E_{2}

is :

View answer and explanation

Correct option: C

Option C (4) is correct because the length of the minor axis of ellipse

E_2

is calculated using the product of eccentricities and the given equation.

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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)

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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