Back to ELLIPSE

JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Application +4 | -1

If

m

is the slope of a common tangent to the curves

\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1

and

\,{{\text{x}}^2} + {y^2} = 12,

then

{\text{12}}{m^2}

is equal to:

View answer and explanation

Correct option: B

9

Option B (9) is correct because after finding the condition for a common tangent and solving for

m^2

, we find that

12m^2 = 9

.

More Questions from ELLIPSE

Practice All Chapter MCQs →

\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

S

S'.

A

be the area of triangle

PSS',

then the maximum value of

A

is If the co-ordinates of two points

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

respectively and

P

is any point on the conic,

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

\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

P

x=0

Q

R

be the image of

Q

y=x

, then circle whose extremities of a diameter are

Q

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

O