JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Analyzing +4 | -1

Let the line

y = mx

and the ellipse

2{x^2} + {y^2} = 1

intersect at a point

P

in the first quadrant. If the normal to this ellipse at

P

meets the co-ordinate axes at

\left( { - \frac{1}{{3\sqrt 2 }},0} \right)

and

\left( {0,\,\beta } \right),

then

\beta

is equal to

View answer and explanation

Correct option: B

\frac{{\sqrt 2 }}{3}

Option B (

\frac{{\sqrt 2 }}{3}

) is correct because by finding the equation of the normal to the ellipse and using the given intercepts, we can determine the value of

\beta

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