Question

Find the equation of a tangent of the hyperbola $4{{x}^{2}}-5{{y}^{2}}=20$ which is parallel to the line $x-y=2$ ?
(a) $x-y+9=0$ ,
(b) $x-y+7=0$ ,
(c) $x-y+1=0$ ,
(d) $x-y-3=0$ .

Answer 1

We start solving the problem by converting the given equation of hyperbola into the standard form. Using this standard form, we write the slope form of the equation of the tangents. We use the fact that the slopes of parallel lines are equal and we find the slope of the given line $x-y=2$ . We substitute the obtained slope in the slope form to get the required equation of the tangents.

Complete step-by-step answer:
According to the problem, we have the equation of the hyperbola given as $4{{x}^{2}}-5{{y}^{2}}=20$ . We need to find the tangent of the given hyperbola which is parallel to the line $x-y=2$ .
Let us draw the given information to get the better view.

We know that the standard form of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Let us convert the given equation of hyperbola $4{{x}^{2}}-5{{y}^{2}}=20$ into the standard form.
$\Rightarrow \dfrac{4{{x}^{2}}-5{{y}^{2}}}{20}=\dfrac{20}{20}$ .
$\Rightarrow \dfrac{4{{x}^{2}}}{20}-\dfrac{5{{y}^{2}}}{20}=1$ .
$\Rightarrow \dfrac{{{x}^{2}}}{5}-\dfrac{{{y}^{2}}}{4}=1$ .
$\Rightarrow \dfrac{{{x}^{2}}}{{{\left( \sqrt{5} \right)}^{2}}}-\dfrac{{{y}^{2}}}{{{2}^{2}}}=1$.
Compared with the standard form, we get ${{a}^{2}}=5$ and ${{b}^{2}}=4$ .
We know that slope equation of the tangent of the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is given by $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$ ---(1).
We substitute the values ${{a}^{2}}=5$ and ${{b}^{2}}=4$ in equation (1) we get the slope form of equation of tangent as $y=mx\pm \sqrt{5{{m}^{2}}-4}$ ---(2).
According to the problem, we need to find the tangent that is parallel to the line $x-y=2$ .
We know that the slopes of parallel lines are equal. So, let us find the slope of the line $x-y=2$ .
We know that slope (m) of the line $ax+by+c=0$ is defined as $m=\dfrac{-a}{b}$ .
So, we get the slope of the line $x-y=2$ is $m=\dfrac{-1}{-1}$ .
$\Rightarrow m=1$ .
Since, the slope of the tangent is equal to the slope of the line $x-y=2$ . We get the slope of the tangent as ‘1’.
Let us substitute ‘1’ in place of m in equation (1).
We get the equation of the tangent as $y=\left( 1 \right)x\pm \sqrt{5{{\left( 1 \right)}^{2}}-4}$ .
$\Rightarrow y=x\pm \sqrt{5\left( 1 \right)-4}$ .
$\Rightarrow y=x\pm \sqrt{5-4}$ .
$\Rightarrow y=x\pm \sqrt{1}$ .
$\Rightarrow y=x\pm 1$ .
$\Rightarrow x-y\pm 1=0$ .
We get the equation of the pair of tangents that is parallel to the line $x-y=2$ as $x-y\pm 1=0$ .
From the options, we can see there is an equation $x-y+1=0$ .
∴ The equation of the tangents parallel to $x-y=2$ as $x-y\pm 1=0$ .

So, the correct answer is “Option c”.

Note: We can see that the options contain only one equation of the tangent. If we are answering the multiple answer type question, we need to see whether there is another equation or not. We can also solve the problem, by taking the equation of the tangent as $x-y=k$ and substituting in the given equation of hyperbola to get the value of k.