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# Derive the equation for the parabola ${y^2} = 4ax$ in standard form.

Last updated date: 23rd Jul 2024
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Hint: Here, we will derive the equation of Parabola in Standard form. We will construct a Parabola with the Coordinates of the focus and the Directrix. We will use the distance between two points formula to find the Distance between the Directrix and the Focus and by equating both the distance where the Directrix is Perpendicular to the $x$- axis and $y$-axis, we will find the equation of the Parabola. Thus, we will derive the equation of Parabola in the standard form.

Formula Used:
We will use the following formulas:
1. Distance between two points is given by the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ where $({x_1},{y_1})$ and $({x_2},{y_2})$ be the two points.
2. The square of the sum of the numbers is given by an algebraic identity ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
3. The square of the difference of the numbers is given by an algebraic identity ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$

Complete Step by Step Solution:
We will find the equation of Parabola in Standard form.
Let $F$ be the focus and $l$ be the Directrix.
Thus, the coordinates of the Focus be$F = \left( {a,0} \right)$ .
Thus, the equation of the Directrix is $x = - a \Rightarrow x + a = 0$
Now, we will draw $FM$ perpendicular to the Directrix $l$. Let the distance between the Directrix and the focus $2a$
$\Rightarrow FM = 2a$
Now, we will take the midpoint of $FM$ as Origin $O$.

So, we get $\left| {OF} \right| = \left| {OM} \right| = a$
Let $P\left( {x,y} \right)$ be any point on the Parabola such that$PF = PB$
We have $PB$ perpendicular to the Directrix$l$ . So, we have the Coordinates of F as $F\left( {a,0} \right)$, M as $M\left( { - a,0} \right)$, B as $B\left( { - a,y} \right)$ .
Distance between two points is given by the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ where $({x_1},{y_1})$ and $({x_2},{y_2})$ be the two points.
Now, by using the Distance formula we will find the distance between $PF$, we get
$\Rightarrow PF = \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - 0} \right)}^2}}$
$\Rightarrow PF = \sqrt {{{\left( {x - a} \right)}^2} + {{\left( y \right)}^2}}$ …………………………………..$\left( 1 \right)$
Now, by using the Distance formula we will find the distance between$PB$ , we get
$\Rightarrow PB = \sqrt {{{\left( {x - \left( { - a} \right)} \right)}^2} + {{\left( {y - y} \right)}^2}}$
$\Rightarrow PB = \sqrt {{{\left( {x + a} \right)}^2}}$ ………………………………………………..$\left( 2 \right)$
Now, by equating the equations $\left( 1 \right)$and$\left( 2 \right)$, we get
$\Rightarrow \sqrt {{{\left( {x - a} \right)}^2} + {y^2}} = \sqrt {{{\left( {x + a} \right)}^2}}$
Now, taking square root on both the sides, we get
$\Rightarrow {\left( {x - a} \right)^2} + {y^2} = {\left( {x + a} \right)^2}$
The square of the sum of the numbers is given by an algebraic identity ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
The square of the difference of the numbers is given by an algebraic identity ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$
Now, by using an algebraic identity, we get
$\Rightarrow {x^2} + {a^2} - 2ax + {y^2} = {x^2} + {a^2} + 2ax$
Now, by cancelling out the terms, we get
$\Rightarrow {y^2} = 2ax + 2ax$
$\Rightarrow {y^2} = 4ax$

Therefore, the equation of Parabola is ${y^2} = 4ax$ in standard form.

Note:
We know that the Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose endpoints lie on the parabola. The Directrix of Parabola is perpendicular to the axis of symmetry and the Directrix does not touch the Parabola. A parabola is symmetric with its axis. If the equation has a ${y^2}$ term, then the axis of symmetry is along the x-axis and if the equation has an ${x^2}$ term, then the axis of symmetry is along the y-axis. So, the given equation of Parabola is open rightwards.