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June 2017 p13 q6
1180
\(The line 3y + x = 25 is a normal to the curve y = x^2 - 5x + k. Find the value of the constant k.\)
Solution
The gradient of the line 3y + x = 25 is \\(-\frac{1}{3}\\). Since this line is normal to the curve, the gradient of the tangent to the curve at the point of intersection is 3.
Differentiate the curve equation y = x^2 - 5x + k to find the gradient of the tangent: \\(\frac{dy}{dx} = 2x - 5\\).
Set the gradient equal to 3: \\(2x - 5 = 3\\).
Solve for \\(x\\):
\\(2x - 5 = 3\\)
\\(2x = 8\\)
\\(x = 4\\)
Substitute \\(x = 4\\) into the line equation to find \\(y\\):
\\(3y + 4 = 25\\)
\\(3y = 21\\)
\\(y = 7\\)
Substitute \\(x = 4\\) and \\(y = 7\\) into the curve equation to find \\(k\\):
