Question

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

Answer 1

Equating the equation of the ellipse with the second-degree equation

$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ with $\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0$

we get $A=\frac{1}{2-r},B=0,C=\frac{1}{r-5},D=0,E=0$ and $F=1$

For the second degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition $B^2-4AC<0$ and also $A\ne C$

$\Rightarrow {\left(0\right)}^2-4\left(\frac{1}{2-r}\right)\left(\frac{1}{r-5}\right)<0$

$\Rightarrow -4\left(\frac{1}{2-r}\right)\left(\frac{1}{r-5}\right)<0$

$\Rightarrow \left(\frac{1}{2-r}\right)\left(\frac{1}{r-5}\right)>0$

$\Rightarrow (2-r)(r-5)<0$

$\Rightarrow (r-2)(r-5)>0$

$\Rightarrow r>2$ and $r>5$