JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

Let the ellipse,

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

, passes through the point

(2,3)

and have eccentricity equal to

\frac{1}{2}

. Then, the equation of the normal to the ellipse at

(2,3)

View answer and explanation

Correct option: B

2 x-y=1

Option B (

2 x-y=1

) is correct because it represents the equation of the normal to the ellipse at (2,3) calculated using the given conditions.

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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