Question

Determine the combined equation of asymptotes for the hyperbola
$\frac{x^2}{25}-\frac{y^2}{16}=1$
If the equation of conjugate hyperbola is $\frac{x^2}{25}-\frac{y^2}{16}=-1$.

• A. $(\frac{x^2}{16}-\frac{y^2}{25})=0$
• B. $(\frac{x^2}{25}-\frac{y^2}{16})=0$
• C. $2(\frac{x^2}{25}-\frac{y^2}{16})=0$
• D. $(\frac{x^2}{25}+\frac{y^2}{16})=0$

Answer 1

The correct option is B $(\frac{x^2}{25}-\frac{y^2}{16})=0$

Given: Hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ ; Conjugate hyperbola is $\frac{x^2}{25}-\frac{y^2}{16}=-1$

To Find: Combined equation of asymptotes

We know that, the sum of equations of hyperbola and its conjugate is twice the equation of asmyptotes.

Hence, adding the given hyperbola equations.

$\Rightarrow (\frac{x^2}{25}-\frac{y^2}{16}-1)+(\frac{x^2}{25}-\frac{y^2}{16}+1)=2\times \text{equation of Asymptotes}$

$\Rightarrow \frac{x^2}{25}-\frac{y^2}{16}+\frac{x^2}{25}-\frac{y^2}{16}=2\times \text{equation of Asymptotes}$

$\Rightarrow 2(\frac{x^2}{25}-\frac{y^2}{16})=2\times \text{equation of Asymptotes}$

$\Rightarrow \text{equation of Asymptotes}:(\frac{x^2}{25}-\frac{y^2}{16})=0$