Tangents and normals to the parabola and rectangular hyperbolaEdexcel International A Level Further Maths: Revision notes

The tangent to a curve at a point touches it there and has the same gradient as the curve. The normal is the line through the same point at right angles to the tangent.

1. Differentiate to get $\frac{dy}{dx}$ and substitute the $x$-coordinate to find the tangent gradient $m$.
2. The normal gradient is $-\frac1m$ (negative reciprocal).
3. Use $y-y_1=m(x-x_1)$ with the point $(x_1,y_1)$. Tangent and normal always pass through the point on the curve.

For $y^2=4ax$ with $y>0$, write $y=2\sqrt a\,x^{1/2}$ and differentiate: $\frac{dy}{dx}=\sqrt a\,x^{-1/2}=\frac{\sqrt a}{\sqrt x}.$ Parametric differentiation is not needed at this level. Example: $y^2=12x$ (so $a=3$) at $(12,12)$: $\frac{dy}{dx}=\frac{\sqrt3}{\sqrt{12}}=\frac12$. Tangent: $y-12=\frac12(x-12)$, i.e. $2y=x+12$. Normal gradient $-2$: $y-12=-2(x-12)$, i.e. $y=-2x+36$. Find the $y$-coordinate of a point from $y^2=4ax$ first when only $x$ is given, and take the sign that the question states.

Write $y=\frac{c^2}{x}=c^2x^{-1}$ and differentiate: $\frac{dy}{dx}=-\frac{c^2}{x^2}.$ The gradient is always negative, so the tangent slopes downwards at every point. Example: $xy=24$ at $A(4,6)$: $\frac{dy}{dx}=-\frac{24}{16}=-\frac32$. Tangent: $y-6=-\frac32(x-4)$, i.e. $3x+2y=24$. Normal gradient $\frac23$: $y-6=\frac23(x-4)$, i.e. $2x-3y+10=0$.

• Axes: put $y=0$ for the $x$-intercept and $x=0$ for the $y$-intercept in the line equation. For the tangent $y=\frac12x+4$ to $y^2=8x$, $y=0$ gives $x=-8$.
• Another line or the curve again: solve the two equations simultaneously. For the normal at $P(2,8)$ to $xy=16$, $y=\frac{x+30}{4}$ gives $x^2+30x-64=0$. One root, $x=2$, is the point you started from, so factorise using it: $(x-2)(x+32)=0$ and the new point has $x=-32$.
• Two tangents: write both equations and solve simultaneously for the intersection.

Unless the question asks for a form, $y=mx+c$ is fine. If it asks for $ax+by+c=0$ with integer coefficients, multiply to clear fractions: $y-6=\frac23(x-4)$ becomes $3y-18=2x-8$, so $2x-3y+10=0$. Check by substituting the point on the curve into your final equation: both sides must agree.