Hint: Whenever you come up with this type of problem firstly convert the given equation to standard form then find the values of ${x_0},{y_0}$ and a. Then check it with the given condition.
Complete step by step answer:
As we know, the standard equation of parabola is $({x - {x_0})^2} = 4a(y - {y_0})$. In which,
$\Rightarrow$ Vertex = $\left( {x{}_0,{y_0}} \right)$
Given Equation of parabola is $y = {x^2} - 8x + c{\text{ }}$
First we have to convert the given equation into the standard form of equation of parabola.
Adding 16 both sides of the given equation to make it a perfect square, it becomes,
$\Rightarrow y + 16 = {x^2} - 8x + 16 + c$
Taking c to LHS of the equation we get,
$\Rightarrow \left( {y + 16 - c} \right) = {\left( {x - 4} \right)^2}{\text{ }}\left( 1 \right)$
Comparing equation 1 with standard equation of parabola we get,
$\Rightarrow {x_0} = 4,{\text{ }}{y_0} = c - 16{\text{ and }}a = \frac{1}{4}$
So, vertex of the equation 1 will be
$\Rightarrow {\text{vertex }} = \left( {4,c - 16} \right)$
According to the question, the vertex lies on the x axis which means that y - coordinate (ordinate) of the vertex should be zero.
So, c - 16 = 0
$\Rightarrow c = 16$
Hence the correct option for the question will be d.
NOTE: - Abscissa is the x - coordinate of a point and ordinate is the y - coordinate of a point.
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