Edexcel IAL - Further Pure Mathematics 1- 4.4 Tangents and Normals to These Curves- Study notes  - New syllabus

Tangents and Normals to the Parabola and Rectangular Hyperbola

The equations of tangents and normals to curves can be found using differentiation. For the syllabus, differentiation is required for the following functions:

\( y = 2a^{\tfrac{1}{2}}x^{\tfrac{1}{2}} \), \( y = \dfrac{c^2}{x} \)

These correspond to the parabola \( y^2 = 4ax \) and the rectangular hyperbola \( xy = c^2 \).

Tangent and Normal to the Parabola

The parabola:

\( y^2 = 4ax \)

can be written as:

\( y = 2\sqrt{a}\sqrt{x} \)

Differentiating with respect to \( x \):

\( \dfrac{dy}{dx} = 2\sqrt{a} \cdot \dfrac{1}{2}x^{-\tfrac{1}{2}} = \sqrt{\dfrac{a}{x}} \)

This gives the gradient of the tangent at any point on the parabola.

Using the Parameter

A general point on the parabola is \( (at^2, 2at) \).

At this point:

\( \dfrac{dy}{dx} = \sqrt{\dfrac{a}{at^2}} = \dfrac{1}{t} \)

So the slope of the tangent is \( \dfrac{1}{t} \).

Equation of the Tangent

At \( (at^2, 2at) \):

\( y – 2at = \dfrac{1}{t}(x – at^2) \)

Equation of the Normal

The slope of the normal is the negative reciprocal:

\( -t \)

So the normal is:

\( y – 2at = -t(x – at^2) \)

Tangent and Normal to the Rectangular Hyperbola

The hyperbola:

\( y = \dfrac{c^2}{x} \)

Differentiating:

\( \dfrac{dy}{dx} = -\dfrac{c^2}{x^2} \)

So the gradient of the tangent at any point is:

\( -\dfrac{c^2}{x^2} \)

Using the Parameter

A general point is \( (ct, \dfrac{c}{t}) \).

At this point:

\( \dfrac{dy}{dx} = -\dfrac{c^2}{c^2t^2} = -\dfrac{1}{t^2} \)

Equation of the Tangent

At \( (ct, \dfrac{c}{t}) \):

\( y – \dfrac{c}{t} = -\dfrac{1}{t^2}(x – ct) \)

Equation of the Normal

The slope of the normal is \( t^2 \).

\( y – \dfrac{c}{t} = t^2(x – ct) \)