Question

If from $\left(1,\alpha \right)$ two tangents are drawn on exactly one branch of the hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$, then ${}^{\prime }{\alpha }^{\prime }$ belongs to

• A. $\left(-1,-\frac{1}{2}\right)$
• B. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• C. $\left(\frac{1}{2},1\right)$
• D. none of these

Answer 1

Answer: (D)

none of these

$Lety=mx+c$ is tangent to the hyperbola

Then tangent in slope form

$\begin{array}{c}\Rightarrow y=mx\pm \sqrt{a^2{m}^2-b^2}\\ \Rightarrow y=mx\pm \sqrt{4m^2-1}\\ passesthrough\left(1,\alpha \right)\\ then,\\ \alpha =m\pm \sqrt{m^2-1}\\ \Rightarrow \left(\alpha -m\right)=\pm \sqrt{4m^2-1}\end{array}$

Squares on both side

$\begin{array}{c}\Rightarrow {\left(\alpha -m\right)}^2=4m^2-1\\ \Rightarrow {\alpha }^2-2\alpha m+m^2=4m^2-1\\ \Rightarrow {\alpha }^2-2\alpha m-3m^2+1=0\\ \Rightarrow 3m^2+2\alpha m-{\alpha }^2-1=0\\ \Rightarrow 3m^2+2\alpha m-\left({\alpha }^3+1\right)=0\\ \Rightarrow D^2-4ac0\\ \Rightarrow 4{\alpha }^2+4\times 3\left({\alpha }^2-1\right)>0\\ \Rightarrow 4{\alpha }^2+12{\alpha }^2-12>0\\ \Rightarrow +16{\alpha }^2-12>0\\ \Rightarrow {\alpha }^2>\frac{12}{16}\end{array}$