JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

A point on the ellipse

\frac{x^2}{16}+\frac{y^2}{9}=1

at a distance equal to the mean of lengths of the semi-major and semi-minor axis from the centre, is

View answer and explanation

Correct option: A

\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)

Option A (

\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)

) is correct because substituting the parametric coordinates into the distance equation and solving yields this point.

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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