Question

# Find the equation of tangents to the hyperbola ${x^2} - 4{y^2} = 4$which are $({\text{i}})$Parallel$({\text{ii}})$Perpendicular to the line $x + 2y = 0.$

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Hint : Since slope of the line is given assume the equation of tangent in slope form and proceed.
The given hyperbola can also be written as $\dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{1} = 1.$
On comparing it with standard equation of hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1.$
We come to know ${a^2} = 4,{b^2} = 1$
We also know slope of the given line $x + 2y = 0{\text{ }}$is $- \dfrac{1}{2}$
(i) When the tangent is parallel to the given line then the
slope of the tangent will be $m = {\text{ }}\dfrac{{ - 1}}{2}$
Then we will apply the condition of tangency in hyperbola which is ${c^2} = {a^2}{m^2} - {b^2}$
On putting the values of $a,b,m$which has been obtained above we get,
${c^2} = 4{\left( {\dfrac{{ - 1}}{2}} \right)^2} - {1^2}$
${\text{ }}c = 1 - 1 = 0$
Therefore the equation will be in the form $y = mx + c$
Then,
$y = - \dfrac{1}{2}x \\ x + 2y = 0 \\$
Above equation is the required equation of tangent.
(ii) When the tangent is perpendicular to the given line $x + 2y = 0$
Then the slope $m$ of the tangent will be
$m{\text{ x}}\;\left( {\dfrac{{ - 1}}{2}} \right) = - 1 \\ m = 2 \\$
Then again applying the condition of tangency of hyperbola we get,
${c^2} = {a^2}{m^2} - {b^2}$

Then putting the value of $a,b,m$ we get,
$c = \pm \sqrt {15}$
Therefore the required equation will now be in the form
$y = mx + c$
On putting the values of $m,c$ we get the equation as
$y = 2x \pm \sqrt {15} .$
Note :- In this question we have just applied the condition of tangency of hyperbola and with the help of given data in question we found slope and the values of a & b then we have applied the condition of parallel and perpendicular .