Question

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents

• A. $circle$
• B. $ellipse$
• C. $hyperbola$
• D. $none$

Answer 1

The correct option is C $hyperbola$

Equating the above equation with the second-degree equation

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

we get $A=\frac{1}{1-k},B=0,C=\frac{1}{1+k},D=0,E=0$ and $F=-1$

$\left(i\right)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition $B^2-4AC=0$ and also $A=C$

$\Rightarrow -4\times \frac{1}{1-k}\times \frac{1}{1+k}=0$

$\Rightarrow \frac{1}{1-k^2}=0$

This case does not exist

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

$\Rightarrow -4\times \frac{1}{1-k}\times \frac{1}{1+k}<0$

$\Rightarrow \frac{1}{1-k^2}>0$

$\Rightarrow 1-k^2<0$

$\Rightarrow -k^2<-1$

$\Rightarrow k^2>1$ does not exist since it is given that $k<1$

$\left(iii\right)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition $B^2-4AC>0$ and also $A\ne C$

$\Rightarrow -4\times \frac{1}{1-k}\times \frac{1}{1+k}>0$

$\Rightarrow \frac{1}{1-k^2}<0$

$\Rightarrow 1-k^2>0$

$\Rightarrow -k^2>-1$

$\Rightarrow k^2<1$

$\therefore k<1$

Hence the above equation represents a hyperbola.