Q97 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2 x^2-y^2=4$ is
- A2
- B$\frac{-2 \pm \sqrt{6}}{2}$
- Correct answerC$-1 \pm \sqrt{6}$
- D$\frac{-2 \pm \sqrt{3}}{2}$
Solution
We know that equation of tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y=m x \pm \sqrt{a^2 m^2-b^2}$
Here, equation of hyperbola is
$2 x^2-y^2=4 \Rightarrow \frac{x^2}{2}-\frac{y^2}{4}=1$
$\therefore$ Equation of tangents are $y=m x \pm \sqrt{2 m^2-4}$
Since, tangents passes through $(1,1)$
$\begin{aligned}
& \therefore 1=m \pm \sqrt{2 m^2-4} \Rightarrow 1-m= \pm \sqrt{2 m^2-4} \\
& \Rightarrow 1+m^2-2 m=2 m^2-4 \Rightarrow m^2+2 m-5=0 \\
& \Rightarrow m=\frac{-2 \pm \sqrt{4+20}}{2}=\frac{-2 \pm \sqrt{24}}{2} \\
& \quad=\frac{-2 \pm 2 \sqrt{6}}{2}=-1 \pm \sqrt{6} .
\end{aligned}$
Hence, the correct answer is option: 'C'.