Tangents & Normals (Edexcel International A Level Further Maths)

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Author

Mark Curtis

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Maths

Tangents & Normals to a Parabola

What is a general point on a parabola?

$(3, 6)$ is a fixed point on the parabola $y^{2} = 12 x$
$(3 t^{2}, 6 t)$ is a general point on the parabola $y^{2} = 12 x$
- It is algebraic
- but it still satisfies the equation
- and it can move along the curve (depending on t)
General points are formed from parametric equations
- A general point on $y^{2} = 4 a x$ is $(a t^{2}, 2 a t)$
You can have two or more distinct general points
- $(3 p^{2}, 6 p)$ and $(3 q^{2}, 6 q)$ are general points P and Q on $y^{2} = 12 x$

How do I find the tangent or normal to a parabola at a general point?

To find the tangent to the parabola $y^{2} = 12 x$ at the point $(3 t^{2}, 6 t)$
- use $y - y_{1} = m (x - x_{1})$
- substitute in $x_{1} = 3 t^{2}$ and $y_{1} = 6 t$
- find $m$ using calculus from $\frac{d y}{d x}$ at $(3 t^{2}, 6 t)$
  - Make $y$ the subject of $y^{2} = 12 x$
  - $y = \sqrt{12 x} = 2 \sqrt{3} x^{\frac{1}{2}}$
  - Ignore ±√ for now
  - $\frac{d y}{d x} = 2 \sqrt{3} \times \frac{1}{2} x^{- \frac{1}{2}} = \sqrt{\frac{3}{x}}$
  - At $x_{1} = 3 t^{2}$ , $\frac{d y}{d x} = \sqrt{\frac{3}{3 t^{2}}} = \frac{1}{t}$
- The tangent is $y - 6 t = \frac{1}{t} (x - 3 t^{2})$
The normal has a perpendicular gradient $- \frac{1}{m}$
- $y - 6 t = - t (x - 3 t^{2})$

How do I use tangents and normals that pass through general points?

All methods in coordinate geometry still work
- However points of intersection will be algebraic in $t$
Questions can involve forming and solving equations in $t$

Exam Tip

If you know parametric and implicit differentiation, they are also acceptable methods to find $\frac{d y}{d x}$ .

Worked Example

The point $P$ with coordinates $(4 t^{2}, 8 t)$ lies on the parabola $y^{2} = 16 x$ .

(a) Use calculus to show that the equation of the normal at $P$ is $y + t x = 4 t (t^{2} + 2)$

The normal at P has the form $y - y_{1} = m (x - x_{1})$
Substitute in $x_{1} = 4 t^{2}$ and $y_{1} = 8 t$

$y - 8 t = m (x - 4 t^{2})$

The gradient of the normal is the negative reciprocal of the gradient of the tangent
Find the gradient of the tangent
For example first make y the subject of the curve
(Use the positive square root)

$y = \sqrt{16 x} = 4 x^{\frac{1}{2}}$

Then differentiate

$\begin{array}{rcl} \frac{d y}{d x} & = & 4 \times \frac{1}{2} x^{- \frac{1}{2}} \\ = & \frac{2}{\sqrt{x}} \end{array}$

Then substitute in $x_{1} = 4 t^{2}$

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{2}{\sqrt{4 t^{2}}} \\ = & \frac{1}{t} \end{array}$

The negative reciprocal is $- t$
Therefore the equation of the normal is

$y - 8 t = - t (x - 4 t^{2})$

Expand and rearrange

$\begin{array}{rcl} y - 8 t & = & - t x + 4 t^{3} \\ y & = & - t x + 4 t^{3} + 8 t \\ y + t x & = & 4 t^{3} + 8 t \end{array}$

Then factorise the right-hand side

$y + t x = 4 t (t^{2} + 2)$

(b) Hence find the coordinates of the three points on the parabola at which the normal passes through the point $(9, 0)$ .

The equation of the normal above must pass through the point $(9, 0)$
Substitute in $x = 9$ and $y = 0$

$0 + 9 t = 4 t (t^{2} + 2)$

This forms an equation in $t$
Rearrange and solve

$\begin{array}{rcl} 9 t & = & 4 t^{3} + 8 t \\ 0 & = & 4 t^{3} - t \\ 0 & = & t (4 t^{2} - 1) \\ 0 & = & t (2 t - 1) (2 t + 1) \end{array}$

$t = 0, t = \frac{1}{2}, t = - \frac{1}{2}$

The question asks for the coordinates of the points on the curve, P, $(4 t^{2}, 8 t)$
Substitute in the three values of $t$

$(0, 0), (1, 4), (1, - 4)$

It helps to imagine the situation visually (below) to check the answers make sense

Tangents & Normals to a Rectangular Hyperbola

What is a general point on a rectangular hyperbola?

$(5, 5)$ is a fixed point on the rectangular hyperbola $x y = 25$
$(5 t, \frac{5}{t})$ is a general point on the rectangular hyperbola $x y = 25$
- It is algebraic
- but it still satisfies the equation
- and it can move along the curve (depending on $t$ )
  - $t \neq 0$
General points are formed from parametric equations
- A general point on $x y = c^{2}$ is $(c t, \frac{c}{t})$
You can have two or more distinct general points
- $(5 p, \frac{5}{p})$ and $(5 q, \frac{5}{q})$ are general points P and Q on $x y = 25$

How do I find the tangent or normal to a rectangular hyperbola at a general point?

To find the tangent to the rectangular hyperbola $x y = 25$ at the point $(5 t, \frac{5}{t})$
- use $y - y_{1} = m (x - x_{1})$
- substitute in $x_{1} = 5 t$ and $y_{1} = \frac{5}{t}$
- find $m$ using calculus from $\frac{d y}{d x}$ at $(5 t, \frac{5}{t})$
  - Make $y$ the subject of $x y = 25$
  - $y = \frac{25}{x} = 25 x^{- 1}$
  - $\frac{d y}{d x} = 25 \times (- 1) x^{- 2} = - \frac{25}{x^{2}}$
  - At $x_{1} = 5 t$ , $\frac{d y}{d x} = - \frac{25}{{(5 t)}^{2}} = - \frac{1}{t^{2}}$
- The tangent is $y - \frac{5}{t} = - \frac{1}{t^{2}} (x - 5 t)$
The normal has a perpendicular gradient $- \frac{1}{m}$
- $y - \frac{5}{t} = t^{2} (x - 5 t)$

How do I use tangents and normals that pass through general points?

All methods in coordinate geometry still work
- However points of intersection will be algebraic in $t$
Questions can involve forming and solving equations in $t$

Exam Tip

If you know parametric and implicit differentiation, they are also acceptable methods to find $\frac{d y}{d x}$ .

Worked Example

The point $P$ with coordinates $(2 p, \frac{2}{p})$ and the point $Q$ with coordinates $(2 q, \frac{2}{q})$ lie on the rectangular hyperbola $x y = 4$ , where $p \neq \pm q$ .

(a) Use calculus to show that the equation of the tangent to the curve at $P$ is $p^{2} y = - x + 4 p$ .

The tangent at P has the form $y - y_{1} = m (x - x_{1})$
Substitute in $x_{1} = 2 p$ and $y_{1} = \frac{2}{p}$

$y - \frac{2}{p} = m (x - 2 p)$

Find the gradient of the tangent, for example, first make y the subject of the curve

$y = \frac{4}{x} = 4 x^{- 1}$

Then differentiate

$\begin{array}{rcl} \frac{d y}{d x} & = & - 4 x^{- 2} \\ = & - \frac{4}{x^{2}} \end{array}$

Then substitute in $x_{1} = 2 p$ and simplify

$\begin{array}{rcl} \frac{d y}{d x} & = & - \frac{4}{{(2 p)}^{2}} \\ = & - \frac{4}{4 p^{2}} \\ = & - \frac{1}{p^{2}} \end{array}$

Therefore the equation of the tangent is

$y - \frac{2}{p} = - \frac{1}{p^{2}} (x - 2 p)$

Multiply both sides by $p^{2}$ then rearrange

$\begin{array}{rcl} p^{2} y - 2 p & = & - (x - 2 p) \\ p^{2} y - 2 p & = & - x + 2 p \end{array}$

Make $p^{2} y$ the subject

$p^{2} y = - x + 4 p$

(b) Hence find the $x$ -coordinate of the point of intersection between the tangent at $P$ and the tangent at $Q$ , giving your answer in fully simplified form.

Finding the tangent at Q requires the exact same steps as above, but with $q$ instead of $p$
You can therefore write down the tangent at Q with no working

$q^{2} y = - x + 4 q$

The tangent at P intersects the tangent at Q

Tangents to a rectangular hyperbola intersecting each other

Make $y$ the subject of each tangent and set them equal to each other

$\frac{- x + 4 p}{p^{2}} = \frac{- x + 4 q}{q^{2}}$

Cross-multiply and expand

$\begin{array}{rcl} q^{2} (- x + 4 p) & = & p^{2} (- x + 4 q) \\ - q^{2} x + 4 p q^{2} & = & - p^{2} x + 4 p^{2} q \end{array}$

Bring the $x$ terms to one side and factorise

$p^{2} x - q^{2} x = 4 p^{2} q - 4 p q^{2} (p^{2} - q^{2}) x = 4 p q (p - q)$

Make $x$ the subject and use the difference of two squares

$\begin{array}{rcl} x & = & \frac{4 p q (p - q)}{p^{2} - q^{2}} \\ x & = & \frac{4 p q (p - q)}{(p - q) (p + q)} \end{array}$

As $p \neq \pm q$ , the brackets will never be zero
This means the $(p - q)$ bracket can be cancelled, giving the final answer

$x = \frac{4 p q}{p + q}$

Note: if $p = q$ then point P is the same as point Q
Also, if $p = - q$ , then the tangents at P and Q are parallel

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Tangents & Normals (Edexcel International A Level Further Maths)

Revision Note

Tangents & Normals to a Parabola

What is a general point on a parabola?

How do I find the tangent or normal to a parabola at a general point?

How do I use tangents and normals that pass through general points?

Exam Tip

Worked Example

Tangents & Normals to a Rectangular Hyperbola

What is a general point on a rectangular hyperbola?

How do I find the tangent or normal to a rectangular hyperbola at a general point?

How do I use tangents and normals that pass through general points?

Exam Tip

Worked Example

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Author: Mark Curtis