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

Find An Equation Of A Parabola That Has Curvature 4 At The Origin 2 1

The main objective of this question is to work out an equation of the parabola given the curvature at the origin. 

A parabola is an equation of the curve in which a point on the curve is equidistant from a fixed point known as a focus and a fixed-line known as directrix.

An essential characteristic of the graph of the parabola is that it has an extreme point called the vertex. If the parabola opens upward, the vertex indicates the lowest point or the minimum value on the graph of a quadratic function, and the vertex represents the highest point or maximum value if the parabola opens downward. In both cases, the vertex serves as a pivot point on the graph. The graph is also symmetric, with the axis of symmetry being a vertical line drawn through the vertex.

Expert Answer

If an equation of the form $f(x)=ax^2$ where $a\neq 0$, the equation of the parabola can be worked out using the formula:

$k(x)=\dfrac{|f”(x)|}{[1+(f'(x))^2]^{3/2}}$                                (1)

Now, differentiating $f(x)$ twice with respect to $x$, we get:

$f'(x)=2ax$  and  $f”(x)=2a$

And substituting these derivative in (1):


$k(x)=\dfrac{2|a|}{[1+4a^2x^2]^{3/2}}$                             (2)

Now, evaluate the curvature at the origin. Substitute $k(0)=4$ in (2):



Since, $k(0)=4$

Therefore,  $2|a|=4$

Hence, $a=2$ or $a=-2$

So the equations of the parabola are:

$f(x)=2x^2$  and  $f(x)=-2x^2$


Given the equation of the parabola $y=x^2-5x+6$, work out the $x$ and $y$ intercepts, the axis of symmetry, and the vertex of the parabola.


The $x-$intercepts are the points on the $x-$axis where the parabola intersects the $x-$axis, and thus their $y$ coordinates are equal to zero. As a result, we must solve the following equation:



Hence, the $x-$intercepts are:

$x=2$ and $x=3$

The $y-$intercepts are the points on the $y-$axis where the parabola intersects the $y-$axis, and thus its $x$ coordinates are equal to zero. So substitute $x=0$ in the given equation:


The $y-$intercept is: $y=6$

Now, the equation of vertex of an up-down facing parabola is of the form:

$y=ax^2+bx+c$                                     (1)

where $x_v=-\dfrac{b}{2a}$

and $a=1,b=-5$  and  $c=6$

Therefore,  $x_v=-\dfrac{(-5)}{2(1)}=\dfrac{5}{2}$

Now, substitute $x_v$ in the given equation to find $y_v$:




Hence, the vertex of parabola is:


Geogebra export

The graph of the given parabola

Images/mathematical drawings are created with GeoGebra.

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