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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $(10, 12)$

Vedclass · Answer

(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

Similar Questions

If the focus of an ellipse is $(-1, -1)$,the equation of its directrix corresponding to this focus is $x + y + 1 = 0$,and its eccentricity is $e = \frac{1}{\sqrt{2}}$,then the length of its major axis is

If $x^{2}+9 y^{2}-4 x+3=0$,where $x, y \in R$,then $x$ and $y$ respectively lie in the intervals:

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :

If the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$,then the length of its latus rectum is

$A$ point $P$ moves such that the sum of its distances from the points $(ae, 0)$ and $(-ae, 0)$ is always $2a$. Find the locus of $P$ (where $0 < e < 1$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module