JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

On the ellipse

2{x^2} + 3{y^2} = 1

the points at which the tangent is parallel to

4x = 3y + 4,

are

View answer and explanation

Correct option: B

\left( { - \frac{2}{{\sqrt {11} }},\frac{1}{{\sqrt {11} }}} \right)

\left( {\frac{2}{{\sqrt {11} }}, - \frac{1}{{\sqrt {11} }}} \right)

Option B (

\left( { - \frac{2}{{\sqrt {11} }},\frac{1}{{\sqrt {11} }}} \right)

\left( {\frac{2}{{\sqrt {11} }}, - \frac{1}{{\sqrt {11} }}} \right)

) is correct because it represents the points on the ellipse where the tangent's slope matches the given line's slope.

More Questions from ELLIPSE

Practice All Chapter MCQs →

\frac{x^2}{144}+\frac{y^2}{81}=1

m

\mathrm{M}

\mathrm{mM}

P

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

S

S'.

A

PSS',

A

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

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

\mathrm{PA}+\mathrm{PB}

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

r

\frac{x^2}{25}+\frac{y^2}{16}=1

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O