KCET · Maths · Hyperbola
The distance between the foci of a hyperbola is 16 and its eccentricity is \(\sqrt{2}\). Its equation is
- A \(\frac{x^2}{4}-\frac{y^2}{9}=1\)
- B \(2 x^2-3 y^2=7\)
- C \(y^2-x^2=32\)
- D \(x^2-y^2=32\)
Answer & Solution
Correct Answer
(D) \(x^2-y^2=32\)
Step-by-step Solution
Detailed explanation
Given,
distance between the foci \(=16\) and
\(\begin{aligned} e & =\sqrt{2} \\ 2 a e & =16 \\ a e & =8\end{aligned}\)
\(\Rightarrow \quad a=\frac{8}{\sqrt{2}}=4 \sqrt{2}\)
So,
\(\begin{aligned} & b^2=a^2\left(e^2-1\right) \\ & b^2=32(2-1)=32\end{aligned}\)
Equation of hyperbola is
\(\begin{aligned} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ & \frac{x^2}{32}-\frac{y^2}{32}=1 \\ & x^2-y^2=32\end{aligned}\)
distance between the foci \(=16\) and
\(\begin{aligned} e & =\sqrt{2} \\ 2 a e & =16 \\ a e & =8\end{aligned}\)
\(\Rightarrow \quad a=\frac{8}{\sqrt{2}}=4 \sqrt{2}\)
So,
\(\begin{aligned} & b^2=a^2\left(e^2-1\right) \\ & b^2=32(2-1)=32\end{aligned}\)
Equation of hyperbola is
\(\begin{aligned} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ & \frac{x^2}{32}-\frac{y^2}{32}=1 \\ & x^2-y^2=32\end{aligned}\)
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