The normal to the curve \(y = 2x^{2} + x - 3\) at the point (2, 7) meets the x- axis at the point P. Find the coordinates of P.
Question 1 Report
The normal to the curve \(y = 2x^{2} + x - 3\) at the point (2, 7) meets the x- axis at the point P. Find the coordinates of P.
Answer Details
Differentiate the curve \(y = 2x^2 + x - 3\):
\[\frac{dy}{dx} = 4x + 1\]At \((2, 7)\) the gradient of the tangent is \(4(2) + 1 = 9\). The normal is perpendicular, so its gradient is:
\[m = -\frac{1}{9}\]Equation of the normal through \((2, 7)\):
\[y - 7 = -\tfrac{1}{9}(x - 2)\]The normal meets the \(x\)-axis where \(y = 0\):
\[-7 = -\tfrac{1}{9}(x - 2) \ \Rightarrow\ 63 = x - 2 \ \Rightarrow\ x = 65\]Therefore \(P = (65,\ 0)\).
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