Q113 / 155 · WBJEE 2019 · Mathematics
Equation of a tangent to the hyperbola $5 x^{2}-y^{2}=5$ and which passes through an external point (2,8) is
- Correct answerA$3 x-y+2=0$
- B$3 x+y-14=0$
- Correct answerC$23 x-3 y-22=0$
- D$3 x-23 y+178=0$
Solution
Given equation of hyperbola can be write as
$\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{5})^{2}}=1$
here, $a=1$ and $b=\sqrt{5}$
Since, $y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$
$y=m x \pm \sqrt{m^{2}-5}$
$\Rightarrow \quad(8-2 m)^{2}=\left(\sqrt{m^{2}-5}\right)^{2}$
$\Rightarrow \quad(8-2 m)^{2}=m^{2}-5$
$\Rightarrow 64+4 m^{2}-32 m=m^{2}-5$
$\Rightarrow 3 m^{2}-32 m+69=0$
$\Rightarrow \quad m=\frac{22}{3}, 3 \quad$ [by solving equation $]$
When $m=\frac{23}{3},$ then equation of tangent
$$
(y-8)=\frac{23}{3}(x-2)
$$
$\Rightarrow \quad 3 y-24=23 x-46$
$\Rightarrow \quad 23 x-3 y-46+24=0$
$\Rightarrow \quad 23 x-3 y-22=0$
When $m=3$, then equation of tangent
$$
\begin{aligned}
& &(y-8) &=3(x-2) \\
\Rightarrow & & y-8 &=3 x-6
\end{aligned}
$$
$\Rightarrow \quad 3 x-y-6+8=0$
$\Rightarrow \quad 3 x-y+2=0$
Hence, option (a) and (c) are correct.
Hence, the correct answer is options: 'A', 'C'.