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# The equation of directrix of the parabola $x^2 - 4x - 8y + 12 = 0$ isa. x = 1b. y = 0 c. x = - 1d. y = - 1

Last updated date: 24th Jul 2024
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Hint: In this type of question is given then firstly we reduce the given equation to the standard form of that conic and then compare ${x_0},{y_0}$ and a with the standard equation of parabola. And then find the required parameter by putting values.

As we know, that standard equation of parabola is $(x - {x_0}^2) = 4a(y - {y_0})$. In which,

$\Rightarrow$ Vertex = $\left( {{x_0},{y_0}} \right){\text{ and,}}$

$\Rightarrow {\text{Equation of directrix of parabola is }}y = {y_0} - a$

Given Equation of parabola is ${x^2} - 4x - 8y + 12 = 0$

First we have to convert given equation to the standard equation of parabola

Taking - 8y + 12 to RHS of the given equation it becomes,

$\Rightarrow {x^2} - 4x = 8y - 12$

Adding 4 both sides of the equation it becomes,

$\Rightarrow \left( {{x^2} - 4x + 4} \right) = 8y - 8$

Taking 8 common in RHS equation becomes,

$\Rightarrow {\left( {x - 2} \right)^2} = 8\left( {y - 1} \right){\text{ }}\left( 1 \right)$

Comparing equation 1 with standard equation of parabola we get,

$\Rightarrow {x_0} = 2,{\text{ }}{y_0} = 1{\text{ and }}a = 2$

So, equation of directrix of the equation 1 will be

$\Rightarrow {\text{directrix}} \Rightarrow y = 1 - 2, \Rightarrow y = - 1$

Hence the correct option for the question will be d.

Note: Understand the diagram properly whenever you are facing these kinds of problems. A better knowledge of formulas will be an added advantage.