JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Analyzing +4 | -1

The area of the triangle inscribed in an ellipse is

2 a b \sin \left(\frac{\beta-\gamma}{2}\right) \sin \left(\frac{\gamma-\alpha}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)

, where

\alpha, \beta

\gamma

are the eccentric angles of the vertices, then the condition that the area of a triangle inscribed in an ellipse may be maximum, is

View answer and explanation

Correct option: B

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

Option B (

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

) is correct because the area of the triangle is maximum when the eccentric angles are equally spaced around the ellipse, which occurs when they are

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

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