Question

The equation of the hyperbola referred to the axis as axes of co-ordinate and whose distance between the foci is $16$ and the eccentricity is $\sqrt{2}$.,is
A.\[{{x}^{2}}-{{y}^{2}}=16\]
B. \[{{x}^{2}}-{{y}^{2}}=32\]
C. \[{{x}^{2}}-2{{y}^{2}}=16\]
D. \[{{y}^{2}}-{{x}^{2}}=32\]

Answer 1

Hint: To solve this question we will use the general equation of hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$ and formula of eccentricity and distance between the foci. Using formula of distance between the foci we will determine the value of $a$ and with the help of the formula of eccentricity we will determine the value of $b$. Then we will substitute the value of $a$ and $b$in the general equation of hyperbola and derive its equation.



Formula Used:Eccentricity of hyperbola: $e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}$.
Distance between the foci: $2ae$.



Complete answer:We are given that the distance between the foci of hyperbola is $16$ and the eccentricity is $\sqrt{2}$ and we have to find the equation of this hyperbola.
We know that the distance between foci can be calculated with the help of formula $2ae$. So we will first determine the value of $a$ by substituting the give value of eccentricity and distance between foci in $2ae$.
$\begin{align}
  & 2ae=16 \\
 & 2a\times \sqrt{2}=16 \\
 & a\sqrt{2}=8 \\
 & a=4\sqrt{2}
\end{align}$
Now we will calculate the value of $b$ using the formula of eccentricity $e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}$.
$\begin{align}
  & \sqrt{2}=\sqrt{1+\frac{{{b}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}} \\
 & \sqrt{2}=\sqrt{1+\frac{{{b}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}}
\end{align}$
On squaring both sides we will get,
$\begin{align}
  & {{\left( \sqrt{2} \right)}^{2}}={{\left( \sqrt{1+\frac{{{b}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}} \right)}^{2}} \\
 & 2=1+\frac{{{b}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}} \\
 & \frac{{{b}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}=1 \\
 & b=4\sqrt{2}
\end{align}$
As we know that the general equation of hyperbola is $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$, we will now substitute the value of $a$ and $b$.
$\begin{align}
  & \frac{{{x}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}-\frac{{{y}^{2}}}{{{\left( 4\sqrt{2} \right)}^{2}}}=1 \\
 & \frac{{{x}^{2}}}{32}-\frac{{{y}^{2}}}{32}=1 \\
 & {{x}^{2}}-{{y}^{2}}=32
\end{align}$

The equation of hyperbola is ${{x}^{2}}-{{y}^{2}}$ when distance between the foci is $16$ and the eccentricity is $\sqrt{2}$.

Option ‘B’ is correct

Note: The eccentricity of a hyperbola defines how curved the conic is and for a hyperbola its value is always greater than one.
There are two focus of hyperbola and both lies on the axis and both are at equal distance from the center.