JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

Locus of the point which divides double ordinate of the ellipse

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

in the ratio

1:2

internally, is

View answer and explanation

Correct option: B

\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1

Option B (

\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1

) is correct because it represents the locus of the point dividing the double ordinate in the given ratio.

More Questions from ELLIPSE

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

m

\mathrm{M}

\mathrm{mM}

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

S

S'.

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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

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y=x

Q

R

N P

Q

P

O