JEE MATHEMATICS MCQ: ELLIPSE

1 Hard

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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 angles

\alpha, \beta, \gamma

, respectively is

\left(x_1, y_1\right)

then

\Sigma \cos \alpha \cos \beta+\Sigma \sin \alpha \sin \beta=

View answer and explanation

Correct option: D

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

Option D (

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

) is correct because it represents the correct relationship between the coordinates of the circumcenter and the eccentric angles of the vertices of the equilateral triangle.

More Questions from ELLIPSE

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Let

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

be the equation of an ellipse. If

m

be the minimum value of the segment of tangent to the ellipse intercepted by the coordinate axes and

\mathrm{M}

be the maximum distance of the centre of the ellipse from the normal, then the value of

\mathrm{mM}

is equal to Let

P

be a variable point on the ellipse

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

with foci at

S

and

S'.

A

be the area of triangle

PSS',

then the maximum value of

A

is If the co-ordinates of two points

A

and

B

are

(\sqrt{7}, 0)

and

(-\sqrt{7}, 0)

respectively and

P

is any point on the conic,

9 x^{2}+16 y^{2}=144

, then

\mathrm{PA}+\mathrm{PB}

is equal to :
The slope of a common tangent to the ellipse

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

and a concentric circle of radius

r

, is A tangent to the ellipse

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

at any point

P

meet the line

x=0

at a point

Q

. Let

R

be the image of

Q

in the line

y=x

, then circle whose extremities of a diameter are

Q

and

R

passes through a fixed point. The fixed point is Any ordinate

N P

of an ellipse meets the auxiliary circle at

Q

. Then, the locus of the intersection of the normal at

P

and

O

is the circle