Question

The value of $p$ such that the vertex of $y = {x^2} + 2px + 13$ is $4$ units above the $x - axis$ is ?
A. $\pm 2$
B. $4$
C. $\pm 3$
D. $5$

Answer 1

Hint : The given equation is an example of a parabola . So , here the vertex is addressing the vertex of a parabola . The vertex of a parabola is the point of intersection of the parabola and its line of symmetry . The vertex of regular parabola is situated at the origin , but here it is given as $4$ , therefore we will compare the term after simplifying the given equation to standard form .

Complete step-by-step answer :
Given : $y = {x^2} + 2px + 13$
The standard form of a parabola is given by :
${\left( {x - h} \right)^2} = a\left( {y - k} \right)$
Here $\left( {h,k} \right)$ represents the vertex of the parabola and $a$ is any constant .
Now we will convert the given equation into standard form , we have
$y = {x^2} + 2px + 13$
Using completing the square method , we will add and subtract ${p^2}$ , we get
$y = {x^2} + 2px + {p^2} - {p^2} + 13$
On solving we get ,
$y = {\left( {x + p} \right)^2} + 13 - {p^2}$
On simplifying we get
$y - \left( {13 - {p^2}} \right) = {\left( {x - \left( { - p} \right)} \right)^2}$
Now comparing the above equation with standard form of the parabola we get ,
$\left( {h,k} \right) = \left( { - p,13 - {p^2}} \right)$ .
Now it is given that the vertex is $4$ units above the $x - axis$ , therefore it is the value of $y$ coordinate .
Now equating coordinates of the $y - axis$ we get ,
$13 - {p^2} = 4$
On simplifying we get ,
${p^2} = 9$
Now taking square root on both sides we get ,
$p = \pm 3$ .
Therefore , option (c) is the correct answer .
So, the correct answer is “Option c”.

Note : In the question it is given that the vertex is $4$ units above the $x - axis$ , which means that it is talking about the $y - axis$ , do not compare the terms of the $x - axis$ . Moreover , in the equation if only one of the terms is of degree $2$ , then it resembles a parabola .