The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by a constant $\alpha$ is $\frac{x^2}{\lambda \sec ^2\left(\frac{\alpha}{2}\right)}+\frac{y^2}{\mu \sec ^2\left(\frac{\alpha}{2}\right)}=1$ then $\lambda+\mu$

JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by a constant

\alpha

\frac{x^2}{\lambda \sec ^2\left(\frac{\alpha}{2}\right)}+\frac{y^2}{\mu \sec ^2\left(\frac{\alpha}{2}\right)}=1

then

\lambda+\mu

View answer and explanation

Correct option: A

a^2+b^2

Option A (

a^2+b^2

) is correct because the locus equation simplifies to the given form, and λ and μ are identified as

a^2

and

b^2

respectively.

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\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