Question

The line $y=x+\lambda$ is tangent to the ellipse $2x^2+3y^2=1$. Then $\lambda$ is

• A. $-2$
• B. $1$
• C. $\sqrt{\frac{5}{6}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (C)

$\sqrt{\frac{5}{6}}$

Substitute $y=x+\lambda$ in the ellipse equation $2x^2+3y^2=1$

$\Rightarrow 2x^2+3(x+\lambda {)}^2=1$

$\Rightarrow 2x^2+3(x^2+2\lambda x+{\lambda }^2)=1$

$\Rightarrow 5x^2+6\lambda x+(3{\lambda }^2-1)=0$

Now for given line to be tangent above above quadratic should have equal roots

$\Rightarrow$ discriminant $=0$

$\Rightarrow (6\lambda {)}^2-4(5\left)\right(3{\lambda }^2-1)=0$

$\Rightarrow 9{\lambda }^2-5(3{\lambda }^2-1)=0$

$\Rightarrow 6{\lambda }^2=5$

$\Rightarrow \lambda =\pm \sqrt{\frac{5}{6}}$