Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is
Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is
$3 x+y-14=0$
$3 x-y+2=0$
$x+y+3=0$
$x-y+6=0$
Solution
Given hyperbola is $5 x^2-y^2=5$ or It can be rewritten as
$\frac{x^2}{1}-\frac{y^2}{5}=1$
Here,
$\therefore$ Equation of tangent is
$\begin{aligned} & y=m x \pm \sqrt{a^2 m^2-b^2} \\ & y=m x \pm \sqrt{1 m^2-5}\end{aligned}$
But point $(2,8)$ lies on it.
$\begin{aligned} & \therefore \quad 8=2 m \pm \sqrt{m^2-5} \\ & \Rightarrow(8-2 m)= \pm \sqrt{m^2-5}\end{aligned}$
On squaring both sides, we get
$\begin{array}{rlrl} & 64+4 m^2-32 m & =m^2-5 \\ \Rightarrow \quad 3 m^2-32 m+69 & =0 \\ \Rightarrow \quad & (3 m-23)(m-3) & =0 \\ \Rightarrow & m & =3, \frac{23}{3}\end{array}$
On putting $m=3$ in Eq. (i), we get
$\begin{aligned} y & =3 x \pm \sqrt{3^2-5}=3 x \pm 2 \\ \Rightarrow & y=3 x+2 \text { and } y=3 x-2\end{aligned}$