Question

Equation of one of the tangents passing through $\left( {2,8} \right)$ to the hyperbola $5{x^2} - {y^2} = 5\;$ is
A. $3x + y - 14 = 0$
B. $3x - y + 2 = 0$
C. $x + y + 3 = 0$
D. $x - y + 6 = 0$

Answer 1

First, we will make the given equation into standard form, since we know that the general equation of tangent is given by $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}}$, we also know that point passes through this equation of tangent. Hence, we will put the point in the equation to find the value of slope m. Once we get m, put m, a and b in the equation of the tangent to get the required equation of one of the tangents passing through $\left( {2,8} \right)$.

Complete step by step solution:
According to the question, the given hyperbola is
$5{x^2} - {y^2} = 5\;$
We know that the Standard form of the hyperbola is
$\dfrac{{{x^2}}}{{{a^2}}}-\dfrac{{{y^2}}}{{{b^2}}} = 1$
To convert $5{x^2} - {y^2} = 5\;$ into standard form, divide it by 5, we get
$\Rightarrow \dfrac{{{x^2}}}{1}-\dfrac{{{y^2}}}{5} = 1$

We know that the general equation of tangent is given by
$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}}$ … (1)
According to the question, $\left( {2,8} \right)$ lies on this equation of the tangent

Hence put $x = 2$ and $y = 8$ in (1), we get
$\Rightarrow 8 = m \times 2 \pm \sqrt {{m^2} - 5}$
On rearranging we get,
$\Rightarrow 8 - 2m = \pm \sqrt {{m^2} - 5}$
On Squaring both sides, we get
$\Rightarrow {(8 - 2m)^2} = {m^2} - 5$
We know that ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , Hence we get
$\Rightarrow {8^2} - 2 \times 8 \times 2m + {(2m)^2} = {m^2} - 5$
On simplification we get,
$\Rightarrow 64 - 32m + 4{m^2} = {m^2} - 5$
On adding like terms, we get,
$\Rightarrow 3{m^2} - 32m + 69 = 0$
By middle term splitting, we get
$\Rightarrow 3{m^2} - 9m - 23m + 69 = 0$
On taking factors common we get,
$\Rightarrow 3m(m - 3) - 23(m - 3) = 0$
On taking $m - 3$ common we get,
$\Rightarrow (3m - 23) \times (m - 3) = 0$
Hence, either $m = 3$ or $m = \dfrac{{23}}{3}$

Since we need to find any one equation, we will first consider $m = 3$ for ease of calculation, hence
$\Rightarrow m = 3$
Now, put $m = 3$ in (1)
$\Rightarrow y = 3x \pm \sqrt {9 - 5}$
On simplification we get,
$\Rightarrow y = 3x \pm \sqrt 4$
On solving the root, we get,
$\Rightarrow y - 3x = \pm 2$
Multiplying $- 1$ throughout, we get
$\Rightarrow - y + 3x = \mp 2$

On rearranging we get,
$\Rightarrow 3x - y \pm 2 = 0$
Hence, the required equations are $3x - y + 2 = 0$ and $3x - y - 2 = 0$

Hence, the final answer is Option B.

Note:
In these questions where any one of the equation is required, we should always first solve the equation which would come from whole numbers, natural numbers, or integers as solutions, we should always avoid fractions for the ease of calculations, for example in the above question, we took $m = 3$ and hence, we were able to quickly solve the question.
Also, in these types of questions, we are expected to know all the general equations of tangents and normal of the 2nd-degree curve given to us. For this question, we particularly need it $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}}$.