KCET · Maths · Hyperbola
The distance between the foci of a hyperbola is \( 16 \) and its eccentricity is \( \sqrt{2} \). Its equation is
- A \( x^{2}-y^{2}=32 \)
- B \( \frac{x^{2}}{4}-\frac{y^{2}}{9}=1 \)
- C \( 2 x^{2}-3 y^{2}=7 \)
- D \( y^{2}-x^{2}=32 \)
Answer & Solution
Correct Answer
(A) \( x^{2}-y^{2}=32 \)
Step-by-step Solution
Detailed explanation
Given that \(e=\sqrt{2}\) and \(q e=16\)
So \(a=4 \sqrt{2}\)
We know that, \(b^{2}=a^{2}\left(e^{2}-1\right)\)
So, \(b^{2}=32(2-1)=32\)
General equation of hyberbola is given by,
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Therefore, \(x^{2}-y^{2}=32\)
So \(a=4 \sqrt{2}\)
We know that, \(b^{2}=a^{2}\left(e^{2}-1\right)\)
So, \(b^{2}=32(2-1)=32\)
General equation of hyberbola is given by,
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Therefore, \(x^{2}-y^{2}=32\)
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