Q43 / 155 · WBJEE 2016 · Mathematics
The line $y=x+\lambda$ is tangent to the ellipse $2 x^{2}+3 y^{2}=1 .$ Then $, \lambda$ is
- A-2
- B1
- Correct answerC$\sqrt{\frac{5}{6}}$
- D$\sqrt{\frac{2}{3}}$
Solution
The equation of the line is $y=x+\lambda$
On comparing it with $y=m x+c,$ we get
$$
m=1 \text { and } c=\lambda
$$
The equation of the ellipse is
$$
2 x^{2}+3 y^{2}=1
$$
or
$$
\frac{x^{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}+\frac{y^{2}}{\left(\frac{1}{\sqrt{3}}\right)^{2}}=1
$$
On comparing it with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ we get
$$
a^{2}=\frac{1}{2} \text { and } b^{2}=\frac{1}{3}
$$
If the line touches the ellipse, then
$$
c^{2}=a^{2} m^{2}+b^{2} \Rightarrow \lambda^{2}=\frac{1}{2} \cdot 1+\frac{1}{3}=\frac{5}{6}
$$
$\Rightarrow \quad \lambda^{2}=\frac{5}{6} \Rightarrow \lambda=\sqrt{\frac{5}{6}}$
Hence, the correct answer is option: 'C'.