Question

The equation $\frac{x^2}{(2-\lambda )}-\frac{y^2}{\lambda -5}-1=0$ represents an ellipse, if

• A. $\lambda >5$
• B. $\lambda <2$
• C. $2<\lambda <5$
• D. $2>\lambda >5$

Answer 1

The correct option is B

$\lambda <2$

Explanation for the correct option:

Compute the range of $\lambda$.

A equation $\frac{x^2}{(2-\lambda )}-\frac{y^2}{\lambda -5}-1=0$ of ellipse is given.

Rewrite the equation as follows:

$\frac{x^2}{(2-\lambda )}+\frac{y^2}{-\lambda +5}=1...1$

We know, the general equation of ellipse is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1...2$.

On comapre equation $1$ and equation $2$. we get,

$2-\lambda =a^2$ and $-\lambda +5=b^2$

Since, $2-\lambda$ and $-\lambda +5$ are the squares of real number.

Therefore,

$$\begin{gathered} 2-\lambda >0 \\ \Rightarrow \lambda <2 \end{gathered}$$

Also,

$$\begin{gathered} -\lambda +5>0 \\ \Rightarrow \lambda <5 \end{gathered}$$

Since, $\lambda <2$ and $\lambda <5$.

So, the intersection of both the inequalities is $\lambda <2$.

Hence, option $B$ is correct .