Question

How do you find the vertex, focus and directrix of ${y^2} + x + y = 0$ ?

Answer 1

Hint: In this question, we are going to find the vertex, focus and directrix of the parabola for the given equation. The given question contains a parabolic equation and we have to compare it to the standard form of parabola. By comparing those we will get the value of vertex, focus and directrix of the parabola. Hence, we can get the required result.

Complete step by step answer:
Given parabolic equation is ${y^2} + x + y = 0$.In order to find the vertex, focus and the directrix of the given parabolic equation, we first have to convert it to the standard form of parabolic equation i.e.,$4p\left( {x - h} \right) = {\left( {y - k} \right)^2}$
The given equation can be rewritten as
$x = - \left( {{y^2} + y} \right) \\
\Rightarrow x = - {\left( {y + \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \\
\Rightarrow \left( {x - \dfrac{1}{4}} \right) = - {\left( {y + \dfrac{1}{2}} \right)^2} \\ $
On comparing both the equations, we get
$h = \dfrac{1}{4} \\
\Rightarrow k = - \dfrac{1}{2} \\
\Rightarrow 4p = 1 \to p = \dfrac{1}{4} \\ $
The vertex is at $\left( {h,k} \right)$.So, the vertex is $\left( {\dfrac{1}{4}, - \dfrac{1}{2}} \right)$, which means that equation represents a horizontal parabola which opens to the left. Since, the focus is $\dfrac{1}{4}$ to the left of the vertex, so it means that Focus is $\left( {0, - \dfrac{1}{2}} \right)$.Moreover, the directrix is $p$ which is to the right of the vertex so,directrix is $x = \dfrac{1}{2}$.

Hence, the vertex is $\left( {\dfrac{1}{4}, - \dfrac{1}{2}} \right)$, the focus is $\left( {0, - \dfrac{1}{2}} \right)$ and the directrix is $x = \dfrac{1}{2}$.

Note:The given question was an easy one. Students should be aware of the parabolas and related concepts. A curve where any point is at an equal distance from a fixed point and a fixed straight line is known as a parabola. Students should be extra careful while comparing the given equation with the standard form of equation.