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# Find the vertex and focus of the parabola and sketch its graph $x +{y^2} = 0$  Hint: To solve these kinds of questions, compare the given equation with the general equation of the curve and find some prominent points like vertex and focus to plot the graph.

The given Equation is $x + {y^2} = 0$

Or we can write it as, ${y^2} = - x$      -- Equation(1)

As you know that general equation of horizontal parabola is

${(y - k)^2} = 4a(x - h)$        -- Equation(2)

Where (h,k) is the coordinate of vertex and a is, distance of focus from vertex.

Now as (h,k) = (0,0) in given equation

So, we can say that vertex is (0,0)

As vertex is origin and parabola is horizontal we can say that focus lies

on x - axis and is equal to (a,0)

And comparing Equation(1) and Equation(2)

We get a = - $\frac{1}{4}$

To plot it's graph we know that its vertex is at (0,0) and it's focus is at $(- \frac{1}{4},0 )$

So, the parabola opens towards the negative x - axis.

Thus the graph would be:

Note: Having a good understanding of conic sections is an added advantage to solve these kinds of problems.

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