Solveeit Logo

Question

Question: If the vertex of the parabola \(y = {x^2}-8x + c\) lies on x - axis, then the value of c is (a) - 1...

If the vertex of the parabola y=x28x+cy = {x^2}-8x + c lies on x - axis, then the value of c is

(a) - 16

(b) - 4

(c) 4

(d) 16

Explanation

Solution

Hint: Whenever you come up with this type of problem firstly convert the given equation to standard form then find the values of x0,y0{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 (xx0)2=4a(yy0)({x - {x_0})^2} = 4a(y - {y_0}). In which,

\Rightarrow Vertex = (x0,y0)\left( {x{}_0,{y_0}} \right)

Given Equation of parabola is y=x28x+c 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,

y+16=x28x+16+c\Rightarrow y + 16 = {x^2} - 8x + 16 + c

Taking c to LHS of the equation we get,

(y+16c)=(x4)2 (1)\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,

x0=4, y0=c16 and a=14\Rightarrow {x_0} = 4,{\text{ }}{y_0} = c - 16{\text{ and }}a = \frac{1}{4}

So, vertex of the equation 1 will be

vertex =(4,c16)\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

c=16\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.