
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
161.1k+ views
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.
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.
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