JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Application +4 | -1

If the tangent at the point

\left( {4\cos \phi ,\frac{{16}}{{\sqrt {11} }}\sin \phi } \right)

to the ellipse

16{x^2} + 11{y^2} = 256

is also a tangent to the circle

{x^2} + {y^2} - 2x = 15

, then the value of

\phi

View answer and explanation

Correct option: C

\pm \frac{\pi }{3}

Option C (

\pm \frac{\pi }{3}

) is correct because it represents the value of

\phi

for which the tangent to the ellipse is also a tangent to the circle, found by equating the distance from the center of the circle to the tangent line with the radius of the circle.

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