Question

How do you write the equation of each parabola in vertex form given vertex (5,12); point (7,15)?

Answer 1

Hint: In the above that has been mentioned above we already have two points that is the vertex of the parabola and also a point which lies on parabola we will use the vertex equation of parabola and then use the value of vertex and the point that has mentioned to find the thickness of the parabola and then we will substitute the value of thickness of parabola in the vertex equation.

Complete step by step answer:
In the above question that has been mentioned we will first write the vertex equation of parabola which is:
$$\begin{array}{rl}& y=a{(x-h)}^2+k\\ & x=a{(y-k)}^2+h\end{array}$$
in this equation the value k is none other than the y coordinate of the vertex and the value h is none other than x coordinate of vertex. As in the question the vertex of the parabola is mentioned we can use its coordinate in the vertex equation of parabola and we will get the equation as:
$$\begin{array}{rl}& \Rightarrow y=a{(x-5)}^2+12\\ & \Rightarrow x=a{(y-12)}^2+5\end{array}$$
so after putting in the vertex value of the equation we can clearly see that the vortex equation of parabola is no more general, now to find the thickness of parabola we will be substituting the point that has been mentioned in the question, this point that has been mentioned in the question will replace x and y in the new formed equations with the help of vertex, the point that has been mentioned in the question lies on the parabola that is why we can use it in the vertex equations and after substituting the point in these equations we will be able to find the thickness of the parabola i.e. “a” value in the equation.
Now the value of “a” in both the equations is as follows
$$\begin{array}{rl}& \Rightarrow 15=a{(7-5)}^2+12\\ & \Rightarrow 3=a{\left(2\right)}^2\\ & \Rightarrow a={\displaystyle \frac{3}{4}}\end{array}$$
$$\begin{array}{rl}& \Rightarrow 7=a{(15-12)}^2+5\\ & \Rightarrow 2=a{\left(3\right)}^2\\ & \Rightarrow a={\displaystyle \frac{2}{9}}\end{array}$$
So the values of a for the two vertex equation are $$a={\displaystyle \frac{3}{4}}$$ and $$a={\displaystyle \frac{2}{9}}$$ respectively.
Now we will substitute the value of a in the respective equation to get the final vertex equation of parabola which will be:
$$\begin{array}{rl}& \Rightarrow y={\displaystyle \frac{3}{4}}{(x-5)}^2+12\\ & \Rightarrow x={\displaystyle \frac{2}{9}}{(y-12)}^2+7\end{array}$$
So the final vertex equation for the vertex and the point mentioned is $$x={\displaystyle \frac{2}{9}}{(y-12)}^2+7$$ and $$y={\displaystyle \frac{3}{4}}{(x-5)}^2+12$$.

Note: In the above question there are some mistakes that can be made which is not using (x-h) and (y-k) in the general equation, remember that this should only be done when the vertex is mentioned as (0,0) if there is some other point on the vertex the equation of the parabola changes too.