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JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

If

y=m x+c

is the tangent on the ellipse

\frac{x^2}{9}+\frac{y^2}{4}=1

, then the value of

c

is

View answer and explanation

Correct option: C

\pm \sqrt{9 m^2+4}

Option C (

\pm \sqrt{{9 m^2+4}}

) is correct because, for the ellipse

\frac{{x^2}}{{9}} + \frac{{y^2}}{{4}} = 1

, the condition for tangency of the line y = mx + c is

c = \pm \sqrt{{a^2m^2 + b^2}} = \pm \sqrt{{9m^2 + 4}}

.

More Questions from ELLIPSE

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\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