Question

Let the line \( y=m x \) 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 \( \left(-\frac{1}{3 \sqrt{2}}, 0\right) \) and \( (0, \beta) \), then \( \beta \) is equal to: 

(A) \( \frac{2}{\sqrt{3}} \) 

(B) \( \frac{2}{3} \) 

(C) \( \frac{2 \sqrt{2}}{3} \) 

(D) \( \frac{\sqrt{2}}{3} \) \( [2020] \)

Answer 1

Correct option is (D) \(\frac{\sqrt 2}3\)

\(2x^2 + (mx)^2 = 1\)

⇒ \(x^2 (2 + m^2) = 1\)

⇒ \(x = \frac 1{\sqrt {2+m^2}}\) in first quadrant

\(\therefore y = \frac m{\sqrt{2 + m^2}}\)

\(P\left(\frac 1{\sqrt{2 + m^2}}, \frac m{\sqrt{2 + m^2}}\right)\).

Normal of ellipse at P is

\(\cfrac {y}{\frac m{\sqrt{2 + m^2}}} - \cfrac x{\frac 2{\sqrt{2 +m^2}}} = 1 - \frac 12 = \frac 12\)

⇒ \(\frac ym - \frac x2 = \frac 1{2\sqrt{1+m^2}}\)

⇒ \(2y - 3x = \frac{m}{\sqrt {2 + m^2}}\)

Put x = 0,

\(2y = \frac m{\sqrt{2 + m^2}}\)

⇒ \(y = \frac m{2\sqrt{2 + m^2}}\)

\(\therefore y = \frac 4{2\sqrt{18}} = \frac 2{3\sqrt 2} = \frac{\sqrt 2}3\)

Put y = 0,

\(-mx = \frac m{\sqrt{2+m^2}}\)

\(x = \frac {-1}{\sqrt{2+m^2}}=\frac{-1}{3\sqrt 2}\)

\(\therefore 2 + m^2 = 18\)

⇒ \(m^2 = 18 - 2\)

⇒ \(m^2 = 16\)

⇒ \(m = \pm 4\)

⇒ \(m = 4\)