The curve represented by frac{x^2}{12-alpha} | vedclass

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(B) The given equation is $\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1$. For this to represent an ellipse,both denominators must be positive. Le

Vedclass · Accepted Answer

an ellipse for all values of $\alpha$ in $(10, 12)$

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(B) The given equation is $\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1$. For this to represent an ellipse,both denominators must be positive. Let $a^2 = 12-\alpha$ and $b^2 = \alpha-10$. For an ellipse,we require $12-\alpha > 0$ and $\alpha-10 > 0$,which implies $10 Since for all $\alpha \in (10, 12)$,both $12-\alpha$ and $\alpha-10$ are positive,the equation represents an ellipse for all values of $\alpha$ in the interval $(10, 12)$. Thus,the correct option is $B$.

The curve represented by $\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1$ is

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