Answer:
B. The value of C is 1
Step-by-step explanation:
Given:
[tex]\displaystyle \large{y=2x+C}\\\displaystyle \large{y^2=8x}[/tex]
To find:
A line being a tangent to curve means both equations are equal but there has to be only one interception between both graphs.
Therefore, substitute y = 2x+C in y² = 8x:
[tex]\displaystyle \large{(2x+C)^2 =8x}[/tex]
Expand the expression (2x+C)² using perfect square:
[tex]\displaystyle \large{4x^2+4xC+C^2=8x}[/tex]
Simplify the equation:
[tex]\displaystyle \large{4x^2+4xC-8x+C^2=0}\\\displaystyle \large{4x^2+(4C-8)x+C^2=0}[/tex]
The discriminant says:
For this, we choose b²-4ac = 0 since tangents only have one intersection.
[tex]\displaystyle \large{b^2-4ac = (4C-8)^2-4(4)C^2 = 0}\\\displaystyle \large{16C^2-64C+64-16C^2=0}\\\displaystyle \large{-64C+64=0}[/tex]
Solve the equation for C:
[tex]\displaystyle \large{-64C=-64}\\\displaystyle \large{C=1}[/tex]
Therefore, the value of C is 1