JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Analyzing +4 | -1

The locus of the centroid of an equilateral triangle inscribed in the ellipse

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

View answer and explanation

Correct option: D

\frac{\left(a^2+3 b^2\right)^2}{a^2} x^2+\frac{\left(b^2+3 a^2\right)^2}{b^2} y^2=\left(a^2-b^2\right)^2

Option D (

\frac{\left(a^2+3 b^2\right)^2}{a^2} x^2+\frac{\left(b^2+3 a^2\right)^2}{b^2} y^2=\left(a^2-b^2\right)^2

) is correct because the locus of the centroid of an equilateral triangle inscribed in the ellipse is given by this equation, derived by using the properties of the centroid and the ellipse equation.

More Questions from ELLIPSE

Practice All Chapter MCQs →

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