Answer
Verified
402.6k+ views
Hint: We will first let the equation of the hyperbola to be \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. Then we will use the fact that the distance between foci of the hyperbola is 2ae and find the value of ‘a’ from it. Then we will use the fact that eccentricity (e) of the hyperbola is $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ to find the value of b.
Complete step-by-step answer:
Now, we have been given that the distance between foci of hyperbola is 16 and its eccentricity is $\sqrt{2}$.
Now, let us take the equation of hyperbola be \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\].
Now, we know that the distance between foci is 2ae. So, using the data given in question we have,
$\begin{align}
& 2ae=16 \\
& 2\times a\times e=16 \\
\end{align}$
Now, we have been given $e=\sqrt{2}$. So, using this we have,
$\begin{align}
& 2a\times \sqrt{2}=16 \\
& a=\dfrac{8}{\sqrt{2}} \\
& a=4\sqrt{2}...........\left( 1 \right) \\
\end{align}$
Now, we know that the eccentricity of hyperbola is,
$e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$
e squaring both sides we have,
$\begin{align}
& {{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
& \left( {{e}^{2}}-1 \right){{a}^{2}}={{b}^{2}} \\
\end{align}$
Now, substituting $e=\sqrt{2}$ and a from (1) we have,
$\begin{align}
& \left( 2-1 \right)16\times 2={{b}^{2}} \\
& 16\times 2={{b}^{2}} \\
& {{b}^{2}}=32..........\left( 2 \right) \\
\end{align}$
Now, we will substituting $a\ and\ {{b}^{2}}$ from (1) and (2) in \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\].
So, we have the equation of hyperbola as \[\dfrac{{{x}^{2}}}{32}-\dfrac{{{y}^{2}}}{32}=1\].
Note: It is important to note that the coordinates of the foci of a hyperbola \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] with e as eccentricity are $\left( ae,o \right),\left( -ae,o \right)$. Also, it is important to note that we have used the fact that eccentricity of hyperbola is $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
Complete step-by-step answer:
Now, we have been given that the distance between foci of hyperbola is 16 and its eccentricity is $\sqrt{2}$.
Now, let us take the equation of hyperbola be \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\].
Now, we know that the distance between foci is 2ae. So, using the data given in question we have,
$\begin{align}
& 2ae=16 \\
& 2\times a\times e=16 \\
\end{align}$
Now, we have been given $e=\sqrt{2}$. So, using this we have,
$\begin{align}
& 2a\times \sqrt{2}=16 \\
& a=\dfrac{8}{\sqrt{2}} \\
& a=4\sqrt{2}...........\left( 1 \right) \\
\end{align}$
Now, we know that the eccentricity of hyperbola is,
$e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$
e squaring both sides we have,
$\begin{align}
& {{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
& \left( {{e}^{2}}-1 \right){{a}^{2}}={{b}^{2}} \\
\end{align}$
Now, substituting $e=\sqrt{2}$ and a from (1) we have,
$\begin{align}
& \left( 2-1 \right)16\times 2={{b}^{2}} \\
& 16\times 2={{b}^{2}} \\
& {{b}^{2}}=32..........\left( 2 \right) \\
\end{align}$
Now, we will substituting $a\ and\ {{b}^{2}}$ from (1) and (2) in \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\].
So, we have the equation of hyperbola as \[\dfrac{{{x}^{2}}}{32}-\dfrac{{{y}^{2}}}{32}=1\].
Note: It is important to note that the coordinates of the foci of a hyperbola \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] with e as eccentricity are $\left( ae,o \right),\left( -ae,o \right)$. Also, it is important to note that we have used the fact that eccentricity of hyperbola is $\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE