# Parabola

## Parabola Definition:

A parabola is the set of points P, whose distance from a fixed point ‘F’ in the plane are equal to their distance from a fixed line ‘l’ in the same plane.

The fixed point F is known as focus and the fixed line is known as directrix of a parabola.

## Terms Related to Parabola:

Let us understand the term related to parabola-

Focus – Focus is defined as a fixed point on the interior of a parabola, which is used to define the curve.

Directrix of a parabola – The Directrix of a parabola is defined as a line that lies behind the parabolic curve. The distance of this line from the vertex is always equal to the distance of the focus from the vertex.

If the focus of a parabolic curve is 5 units away from the vertex, then the Directrix will also be 5 units away from the vertex.

Axis of symmetry of a parabola – A line passing through the focus and perpendicular to the directrix of the parabola, is known as the axis of symmetry of the parabola.

Latus Rectum – A line segment passing through a focus of a parabola having starting and ending points both lying on the parabolic curve is known as Latus Rectum.

## Shape of Parabola –

Depending upon the value of “a” the parabolic shapes are as follows-

• If a > 0, the parabola opens upward and the lowest point (called the vertex) yields the minimum value of the function.
• If |a|<1, the parabola opens downward and the highest point (called the vertex) yields the maximum value of the function.
• If |a|>1, the parabola is narrower than the standard parabola y=x2$y = x^{2}$ but if a < 1 then, it is wider than the standard parabola.

## General form of Parabolic equation-

The standard form of a parabolic curve is derived from the quadratic equation. The quadratic equation is y=ax2+bx+c$y = ax^{2} + bx + c$,

where, a,b,cR$a,b,c \in \mathbb{R}$ and a0$a \neq 0$.

Using the method of completing the square

f(x)=a(x+b2a)+4acb24a$f(x) = a\left ( x + \frac{b}{2a} \right ) + \frac{4ac – b^{2}}{4a}$

which is the equation of a parabola

x=b2a$x = -\frac{b}{2a}$ (that is parallel to y-axis)

focal length = 14a$\frac{1}{4a}$

Vertex (V) = (b2a,4acb24a)$\left ( -\frac{b}{2a}, \frac{4ac – b^{2}}{4a} \right )$

Focus (F) = (b2a,4acb2+14a)$\left ( -\frac{b}{2a}, \frac{4ac – b^{2} + 1}{4a} \right )$

Directrix, y = 4acb2+14a$\frac{4ac – b^{2} + 1}{4a}$

## Standard equation of a Parabola:

The standard form of a parabola is simplest if the vertex lies at the origin. The axis of symmetry lies along the x-axis or y-axis depending upon the orientation of the parabola. The following four types are as follows:

 Equation of Parabola y2=4ax$y^{2} = 4ax$ y2=−4ax$y^{2} = -4ax$ x2=4ay$x^{2} = 4ay$ x2=−4ay$x^{2} = -4ay$ Axis y = 0 y = 0 x = 0 x = 0 Directrix x=−a$x = -a$ x = 0 y=−a$y = -a$ y = 0 Vertex (0,0) (0,0) (0,0) (0,0) Focus (a,0) (−a,0)$(-a,0)$ (0,a) (0,−a)$(0,-a)$ Length of Latus Rectum 4a 4a 4a 4a Equation of Latus Rectum x = a x=−a$x = -a$ y = a y=−a$y = -a$

Example- Find the focus & length of latus rectum of a parabola having an equation –

\(y^{2} = 10x[/latex]

Solution- Given equation of a parabola = y2=10x$y^{2} = 10x$

y2=4(104)x$y^{2} = 4\left (\frac{10}{4} \right )x$

So a=104=52$a = \frac{10}{4} = \frac{5}{2}$

Focus (F) = (a,0)=(52,0)$(a,0) = \left (\frac{5}{2} , 0 \right )$

Length of Latus Rectum = 4a=4(104)=10$4a = 4\left (\frac{10}{4} \right ) = 10$

## Determining the Focal Distance of a point-

Consider a parabola of the form x2=4ay$x^{2} = 4ay$, and point P(x,y) lying on the curve, then distance of the point from the focus (0,a) of a parabola will be given as-

Focal distance (FP) =x2+(ya)2$= \sqrt{x^{2} + (y-a)^{2}}$,

we know, x2=4ay$x^{2} = 4ay$

=4ay+(ya)2$= \sqrt{4ay + (y-a)^{2}}$

=(y+a)2$= \sqrt{(y + a)^{2}}$

Thus, FP=|y+a|$FP = \left | y + a \right |$

The following observation can be observed from the standard equation of the parabola-

(i) The parabola is a symmetric curve, w.r.t. the axis of the parabola. If the equation has a x2$x^{2}$ term, then the axis of symmetry is along y-axis, and vice-versa for the y2$y^{2}$ term.

(ii) If the axis of symmetry is along x-axis, i.e. having a y2$y^{2}$ term, then parabola opens to the –

• Right- If the coefficient of x is positive.
• Left- If the coefficient of x is negative.

(iii) If the axis of symmetry is along y-axis, i.e. having a x2$x^{2}$ term, then parabola opens to the –

• Upwards- If the coefficient of y is positive.
• Downwards- If the coefficient of y is negative.

## Area of a parabola:

### Method 1-

Area of a parabolic arc is given as

23ab$\frac{2}{3}ab$,

where a= height of the parabolic arc

b = base of the parabolic arc

### Method 2- Area using Integration

We can also use the method of integration for calculating the area of a parabola.

### Area of a parabola by Integration:

Let us try to find the area of a shaded region of a parabola

The curve is of the form x=y2$x = y^{2}$ ————————(i)

The shaded region is upto (x,y)=(1,0)$(x,y) = (1,0)$

Area of a shaded region is given as-

Area = 10y.dx$\int_{0}^{1} y.dx$ ( the definite integral is from 0 to 1, as the vertex lies at origin (0) and goes upto 1) ————————(ii)

Substituting the value of y from (i) in equation (ii), we have

y=x12$y = x^{\frac{1}{2}}$

10y.dx=10x12.dx$\Rightarrow \int_{0}^{1} y.dx = \int_{0}^{1} x^{\frac{1}{2}}.dx$

=[23×x32]10$= \left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]^{1}_{0}$

=23×10=23$= \frac{2}{3} \times 1 – 0 = \frac{2}{3}$

Example- Find the area of a parabola for the given function- f(x)=x2+4$f(x) = -x^{2} + 4$

Solution-

baf(x).dx=22(x2+4).dx$\int_{a}^{b}f(x).dx = \int_{-2}^{2} (-x^{2} + 4).dx$

=22(13x3+4x)$= \int_{-2}^{2} \left ( -\frac{1}{3} x^{3} + 4x \right )$

=[x33+4x]22$= \left [ \frac{-x^{3}}{3} + 4x \right ]_{-2}^{2}$

=[13(2)3+4(2)][13(2)3+4(2)]$= \left [ -\frac{1}{3} (2)^{3} + 4(2) \right ] – \left [ -\frac{1}{3} (-2)^{3} + 4(-2) \right ]$

=(83+8)(838)$= (-\frac{8}{3} + 8) – \left ( \frac{8}{3} – 8 \right )$

=(163)(163)$= \left ( \frac{16}{3} \right ) – \left ( – \frac{16}{3} \right )$

=323$= \frac{32}{3}$<

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#### Practise This Question

In the figure given below, identify the exterior angles.