Find normal line equation

Use the negative reciprocal of the gradient at a point to find the equation of the normal line.

3 questions · Moderate -0.1

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CAIE P1 2017 November Q10
11 marks Moderate -0.3
10
\includegraphics[max width=\textwidth, alt={}, center]{518bb805-5b14-4b41-94fd-38a31a90c218-18_551_689_260_726} The diagram shows part of the curve \(y = \sqrt { } ( 5 x - 1 )\) and the normal to the curve at the point \(P ( 2,3 )\). This normal meets the \(x\)-axis at \(Q\).
  1. Find the equation of the normal at \(P\).
  2. Find, showing all necessary working, the area of the shaded region.
Edexcel C3 2006 January Q3
5 marks Moderate -0.3
3. The point \(P\) lies on the curve with equation \(y = \ln \left( \frac { 1 } { 3 } x \right)\). The \(x\)-coordinate of \(P\) is 3 . Find an equation of the normal to the curve at the point \(P\) in the form \(y = a x + b\), where \(a\) and \(b\) are constants.
(5)
OCR C3 Q8
13 marks Standard +0.3
8. The curve \(C\) has the equation \(y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0\).
  1. Find an equation for the normal to the curve at the point \(\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)\). The curve \(C\) has a stationary point with \(x\)-coordinate \(\alpha\) where \(0.5 < \alpha < 1\).
  2. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
  3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$ with \(x _ { 0 } = 1\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the value of \(x _ { 4 }\) to 3 decimal places.
  4. Show that your value for \(x _ { 4 }\) is the value of \(\alpha\) correct to 3 decimal places.
  5. Another attempt to find \(\alpha\) is made using the iterative formula $$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$ with \(x _ { 0 } = 1\). Describe the outcome of this attempt.