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$({\text{i}})$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 .

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 .

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