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

The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

Find the eccentricity of an ellipse,if the length of its latus rectum is $4$ units and the distance between its vertex and the nearest focus is $3/2$ units.

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

The coordinates of the foci of the ellipse $3x^2 + 4y^2 - 12x - 8y + 4 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module