JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

Tangent at a point on the ellipse

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

is drawn which cuts the coordinates axes at

A

and

B

. The minimum area of the triangle

O A B

is (

O

being origin)

View answer and explanation

Correct option: A

a b

Option D (

\left(\frac{4 a m^2}{l^2}, \frac{4 a m}{l}\right)

) is correct because it represents the coordinates of point D, which is the intersection of the third normal from point C with the parabola.

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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