JEE Advanced · Mathematics · 15. Hyperbola
The line \(2 x+y=1\) is tangent to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If this line passes through the point of intersection of the nearest directrix and the \(x\)-axis, then the eccentricity of the hyperbola is
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
On substituting \(\left(\frac{a}{e}, 0\right)\) in \(y=-2 x+1\), we get
\[
\begin{aligned}
0 & =-\frac{2 a}{e}+1 \\
\Rightarrow \quad \frac{a}{e} & =\frac{1}{2}
\end{aligned}
\]
Also, \(y=-2 x+1\) is tangent to hyperbola
\[
\begin{array}{ll}
\therefore & 1=4 a^2-b^2 \\
\Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right)
\end{array}
\]

\[
\begin{array}{ll}
\Rightarrow & \frac{4}{e^2}=5-e^2 \\
\Rightarrow & e^4-5 e^2+4=0 \\
\Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0 \\
\Rightarrow & e=2, e=1
\end{array}
\]
\(e=1\) gives the conic as parabola. But conic is given as hyperbola, hence \(e=2\)
\[
\begin{aligned}
0 & =-\frac{2 a}{e}+1 \\
\Rightarrow \quad \frac{a}{e} & =\frac{1}{2}
\end{aligned}
\]
Also, \(y=-2 x+1\) is tangent to hyperbola
\[
\begin{array}{ll}
\therefore & 1=4 a^2-b^2 \\
\Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right)
\end{array}
\]

\[
\begin{array}{ll}
\Rightarrow & \frac{4}{e^2}=5-e^2 \\
\Rightarrow & e^4-5 e^2+4=0 \\
\Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0 \\
\Rightarrow & e=2, e=1
\end{array}
\]
\(e=1\) gives the conic as parabola. But conic is given as hyperbola, hence \(e=2\)
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