KCET · Maths · Complex Number
If the eccentricity of the hyperbola \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) is \( \frac{5}{4} \) and \( 2 x+3 y-6=0 \) is a focal chord of
the hyperbola, then the length of transverse axis is equal to
- A \( \frac{12}{5} \)
- B \( \frac{24}{5} \)
- C \( \frac{6}{5} \)
- D \( \frac{5}{24} \)
Answer & Solution
Correct Answer
(B) \( \frac{24}{5} \)
Step-by-step Solution
Detailed explanation
Given equation of hyperbola,
\[
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \rightarrow(1)
\]
and a focal chord of the hyperbola is,
\[
2 x+3 y-6=0 \rightarrow(2)
\]
We know that, focus is given by \( (a e, 0) . \) So, from Eq. (2) we have
\[
2 a e-6=0 \Rightarrow a e=3 \Rightarrow a=\frac{3}{e}
\]
\[
\begin{array}{l}
\text { Hence } e=\frac{5}{4} \text { then, } \\
\Rightarrow a=\frac{3}{5 / 4}=\frac{12}{5}
\end{array}
\]
Length of the transverse axis is \( 2 \mathrm{a} . \) So, \( 2 a=\frac{24}{5} \)
\[
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \rightarrow(1)
\]
and a focal chord of the hyperbola is,
\[
2 x+3 y-6=0 \rightarrow(2)
\]
We know that, focus is given by \( (a e, 0) . \) So, from Eq. (2) we have
\[
2 a e-6=0 \Rightarrow a e=3 \Rightarrow a=\frac{3}{e}
\]
\[
\begin{array}{l}
\text { Hence } e=\frac{5}{4} \text { then, } \\
\Rightarrow a=\frac{3}{5 / 4}=\frac{12}{5}
\end{array}
\]
Length of the transverse axis is \( 2 \mathrm{a} . \) So, \( 2 a=\frac{24}{5} \)
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