JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

Consider the particle travelling clockwise on the elliptical path

\frac{x^2}{100}+\frac{y^2}{25}=1

. The particle leaves the orbit at the point

(-8,3)

and travels in a straight line tangent to the ellipse, then the point will the particle cross the

Y

-axis is

View answer and explanation

Correct option: C

\left(0, \frac{25}{3}\right)

Option C ((0,

\frac{25}{3}

)) is correct because the tangent line equation is found and then the y-intercept is calculated by setting x=0.

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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