KCET · Maths · Hyperbola
The equation of a hyperbola whose asymptotes are \(3 x \pm 5 y=0\) and vertices are \((\pm 5,0)\) is
- A \(3 x^{2}-5 y^{2}=25\)
- B \(5 x^{2}-3 y^{2}=225\)
- C \(25 x^{2}-9 y^{2}=225\)
- D \(9 x^{2}-25 y^{2}=225\)
Answer & Solution
Correct Answer
(D) \(9 x^{2}-25 y^{2}=225\)
Step-by-step Solution
Detailed explanation
The given equation of asymptotes of hyperbola is
\(\begin{array}{cc}3 x \pm 5 y=0 \\ \text { and vertices } & (\pm 5,0)\end{array}\)
Now, equation of hyperbola is
\[
\begin{gathered}
(3 x+5 y)(3 x-5 y)=\lambda \\
9 x^{2}-25 y^{2}=\lambda
\end{gathered}
\]
Since, \((\pm 5,0)\) is the vertices of the hyperbola, then
\[
9(5)^{2}-0=\lambda \Rightarrow \lambda=225
\]
From Eq. (ii), \(9 x^{2}-25 y^{2}=225\)
\(\begin{array}{cc}3 x \pm 5 y=0 \\ \text { and vertices } & (\pm 5,0)\end{array}\)
Now, equation of hyperbola is
\[
\begin{gathered}
(3 x+5 y)(3 x-5 y)=\lambda \\
9 x^{2}-25 y^{2}=\lambda
\end{gathered}
\]
Since, \((\pm 5,0)\) is the vertices of the hyperbola, then
\[
9(5)^{2}-0=\lambda \Rightarrow \lambda=225
\]
From Eq. (ii), \(9 x^{2}-25 y^{2}=225\)
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