JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

If circumcentre of an equilateral triangle inscribed in

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

, with vertices having eccentric angle

\alpha, \beta, \gamma

respectively is

(h, k)

, then

\Sigma \cos (\alpha-\beta)=

View answer and explanation

Correct option: D

\frac{9 h^2}{2 a^2}+\frac{9 k^2}{2 b^2}-\frac{3}{2}

Option D (

\frac{{9 h^2}}{{2 a^2}}+\frac{{9 k^2}}{{2 b^2}}-\frac{{3}}{{2}}

) is correct because it relates the circumcenter coordinates (h, k) to the eccentric angles and the ellipse parameters.

More Questions from ELLIPSE

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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'.

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A

B

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r

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

P

x=0

Q

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Q

y=x

Q

R

N P

Q

P

O