JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Analyzing +4 | -1

\tan \theta_1 \tan \theta_2=-\frac{a^2}{b^2}

, then the chord joining two points

\theta_1

and

\theta_2

on the ellipse

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

will subtend a right angle at

View answer and explanation

Correct option: B

centre

The correct answer is B (centre) because the condition

\tan \theta_1 \tan \theta_2=-\frac{a^2}{b^2}

implies that the lines joining the points to the center are perpendicular.

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