JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

Let

P

be a point in the first quadrant lying on the ellipse

9 x^2+16 y^2=144

, such that the tangent at

P

to the ellipse is inclined at an angle of

135^{\circ}

to the positive direction of

x

-axis. Then the coordinates of

P

are

View answer and explanation

Correct option: A

\left(\frac{16}{5}, \frac{9}{5}\right)

Option A ((16/5, 9/5)) is correct because it satisfies the ellipse equation and the tangent at that point has a slope of -1.

More Questions from 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