# Find an equation of a parabola that has curvature 4 at the origin

Here in this question, we have to find the parabola equation, which has a curvature of 4 and it lies at the origin.

As we know that the general equation of the parabola in terms of x-axis and y-axis is given as $y=\ a\ {(\ x – h\ )}^2+\ k$ (regular parabola) or $x=\ a\ {(\ y-k\ )}^2+\ h$ (sideways parabola) where $(h,k)$ are the vertex of parabola.

As given in the question, the parabola lies on the origin so $(h,k)=(0,0)$, now putting this value in the general equation of the parabola we get,

$y=\ a\ {(\ x – 0\ )}^2+\ 0 , ( h, k) = ( 0, 0)$

$y=\ a\ { x }^2+\ 0$

Taking the derivative, we get:

$\frac {dy}{dx}\ =\ \frac {d}{dx}\ , ( a\ x^2 + \ 0 )\ \$

Then our required equation will be,

$f(x) \ =\ a x^2,\ a\neq0$

Now to calculate the curvature we have its formula shown below

$k\ =\ \frac {\left|\ \ \ f^{\prime\prime} \left ( x \right ) \right | } { \left [\ 1\ +\ \left (f^\prime \left ( x \right )\right)^2\ \ \right]^\frac { 3 } { 2 } }$

For this we have to find $f^{\prime\prime} \left ( x \right )$ and $f^\prime \left ( x \right )$

$f^\prime \left ( x \right ) =2ax$

$f^{\prime\prime} \left ( x \right ) =2a$

Putting the values of these differentials in the above formula of curvature

$k\ =\ \frac { \left| \ 2 a\ \right| } { \left[ \ 1\ +\ \left(\ 2\ a\ x\ \right )^2 \ \ \right ]^\frac {3}{2} }$

To find the value of a, evaluate the curvature $k$ at the origin and set $k(0)=4$

we get

$k(0) = 2\left| a\right|=4$

$\left| a\right| = \frac {4}{2}$

The value of a comes out to be $a=2$ or $a=-2$

Putting the values of $a$ in the equation of parabola we have,

$f\left ( x\right) = 2 x^2 ; f\left( x \right) = – 2 x^2$

## Numerical Results:

The required equation of the parabolas are as follows

$f\left(x\right)=2x^2$

$f\left(x\right)=-2 x^2$

## Example:

Equation of a parabola is $y^2=24x$ . Find length of the latus rectum, vertex and focus for given parabola.

Given as,

Equation of parabola: $y^2=24x$

we conclude that $4a=24$

$a= \dfrac{24}{4}=6$

Required parameters are,

Length of latus rectum = $4a=4(6)=24$

Focus = $(a,0)=(6,0)$

Vertex = $(0,0)$

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