Q82 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
The equation $\dfrac{x^2}{2-\lambda}+\dfrac{y^2}{\lambda-5}+1=0$ represents an ellipse, if
- A$\lambda <5$
- B$\lambda <2$ or $\lambda>5$
- Correct answerC$2 <\lambda <5$
- D$\lambda <2$
Solution
The given equation is $\dfrac{x^2}{2-\lambda} + \dfrac{y^2}{\lambda-5} + 1 = 0$.
Rearranging the equation, we get $\dfrac{x^2}{\lambda-2} + \dfrac{y^2}{5-\lambda} = 1$.
For this equation to represent an ellipse, the denominators must be positive and distinct. Thus, we require $\lambda-2 > 0$ and $5-\lambda > 0$.
From $\lambda-2 > 0$, we have $\lambda > 2$.
From $5-\lambda > 0$, we have $\lambda < 5$.
Combining these inequalities, we get $2 < \lambda < 5$.
Additionally, for the equation to represent an ellipse, the denominators must not be equal, i.e., $\lambda-2 \neq 5-\lambda$, which implies $2\lambda \neq 7$ or $\lambda \neq 3.5$. Since $3.5$ is within the interval $(2, 5)$, the condition $2 < \lambda < 5$ is the standard requirement for the general form of an ellipse.
Answer: $2 <\lambda <5$
Rearranging the equation, we get $\dfrac{x^2}{\lambda-2} + \dfrac{y^2}{5-\lambda} = 1$.
For this equation to represent an ellipse, the denominators must be positive and distinct. Thus, we require $\lambda-2 > 0$ and $5-\lambda > 0$.
From $\lambda-2 > 0$, we have $\lambda > 2$.
From $5-\lambda > 0$, we have $\lambda < 5$.
Combining these inequalities, we get $2 < \lambda < 5$.
Additionally, for the equation to represent an ellipse, the denominators must not be equal, i.e., $\lambda-2 \neq 5-\lambda$, which implies $2\lambda \neq 7$ or $\lambda \neq 3.5$. Since $3.5$ is within the interval $(2, 5)$, the condition $2 < \lambda < 5$ is the standard requirement for the general form of an ellipse.
Answer: $2 <\lambda <5$
Hence, the correct answer is option: 'C'.