JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Ellipse

Let the line

y = mx

and the ellipse

2 x^{2} + y^{2} = 1

intersect at a point

P

in the first quadrant. If the normal to this ellipse at P meets the co - ordinate axes at

(- \frac{1}{3 2}, 0)

and

(0, β)

then

β

is equal to:

Answer: (D)

Solution

Let

P (x_{1}, y_{1})

lie on the ellipse

2 x^{2} + y^{2} = 1

, that is

\frac{x ^{2}}{1/2} + \frac{y ^{2}}{1} = 1

. The normal at

P

\frac{x}{2 x _{1}} - \frac{y}{y _{1}} = - \frac{1}{2}

, and since it passes through

(- \frac{1}{3 2}, 0)

\frac{- 1/ ( 3 2 )}{2 x _{1}} = - \frac{1}{2}

, giving

x_{1} = \frac{1}{3 2}

. The ellipse then gives

2 \cdot \frac{1}{18} + y_{1}^{2} = 1

, so

y_{1}^{2} = \frac{8}{9}

and

y_{1} = \frac{2 2}{3}

. Setting

x = 0

in the normal gives the

y

-intercept

y = \frac{y _{1}}{2} = \frac{2}{3}

, so

β = \frac{2}{3}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option