Q72 / 180 · KCET 2015 · Mathematics
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
the hyperbola, then the length of transverse axis is equal to
- A\( \frac{12}{5} \)
- Correct answerB\( \frac{24}{5} \)
- C\( \frac{6}{5} \)
- D\( \frac{5}{24} \)
Solution
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, \( 2 a=\frac{24}{5} \)
Hence, the correct answer is option: 'B'.