JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Analyzing +4 | -1

The tangent at a point

P(\theta)

to the ellipse

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

cuts the auxiliary circle at points

Q

and

R

. If

Q R

subtends a right angle at the centre

C

of the ellipse, then the eccentricity of the ellipse, is

View answer and explanation

Correct option: B

\frac{1}{\sqrt{1+\sin ^2 \theta}}

Option C (An ellipse) is correct because the locus of the point of intersection of tangents at P and Q to the ellipse is another ellipse.

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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