JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

The locus of poles of tangents to the ellipse

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

with respect to concentric ellipse

\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1

, is

View answer and explanation

Correct option: D

\frac{a^2 x^2}{\alpha^4}+\frac{b^2 y^2}{\beta^4}=1

Option D (

\frac{{a^2 x^2}}{{\alpha^4}} + \frac{{b^2 y^2}}{{\beta^4}} = 1

) is correct because the locus of the poles is found by equating the polar equation to the tangent equation and using the condition for tangency.

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