Tangents, normals and lociEdexcel International A Level Further Maths: Revision notes

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, differentiate implicitly to find the gradient at $(x_1,y_1)$. Ellipse: $\frac{dy}{dx}=-\frac{b^2x_1}{a^2y_1}$. Hyperbola: $\frac{dy}{dx}=\frac{b^2x_1}{a^2y_1}$. The tangent has gradient $m$; the normal is perpendicular, with gradient $-\frac1m$. Substituting into $y-y_1=m(x-x_1)$ gives the compact tangent forms $\text{ellipse: }\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1,\qquad\text{hyperbola: }\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1.$ Example: on $\frac{x^2}{25}+\frac{y^2}{9}=1$ at $\left(4,\frac95\right)$ the tangent is $\frac{4x}{25}+\frac{y}{5}=1$, i.e. $4x+5y=25$, with gradient $-\frac45$; the normal has gradient $\frac54$.

An ellipse can be written as $(a\cos\theta,\,b\sin\theta)$ and a hyperbola as $(a\sec\theta,\,b\tan\theta)$. Then $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}$. Ellipse: $\frac{dy}{dx}=-\frac{b\cos\theta}{a\sin\theta}$, tangent $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$, normal $ax\sec\theta-by\operatorname{cosec}\theta=a^2-b^2$. Hyperbola: $\frac{dy}{dx}=\frac{b}{a\sin\theta}$, tangent $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$. The tangent forms come from the Cartesian one by putting $x_1=a\cos\theta$, $y_1=b\sin\theta$ (or $a\sec\theta$, $b\tan\theta$). For a normal, find the gradient first, then use $y-y_1=m(x-x_1)$.

Substitute $y=mx+c$ into the curve to obtain a quadratic in $x$. The line is a tangent when this has a repeated root, so the discriminant is zero. For the ellipse, $b^2x^2+a^2(mx+c)^2=a^2b^2$ gives $(b^2+a^2m^2)x^2+2a^2mc\,x+a^2(c^2-b^2)=0$. Setting the discriminant to zero simplifies to the results $\text{ellipse: }c^2=a^2m^2+b^2,\qquad\text{hyperbola: }c^2=a^2m^2-b^2.$ For the hyperbola a real tangent with gradient $m$ exists only if $a^2m^2\geq b^2$. Example: $\frac{x^2}{16}-\frac{y^2}{9}=1$ and $y=2x+c$: $c^2=64-9=55$, so $c=\pm\sqrt{55}$. With $m=\frac12$, $c^2=4-9<0$, so there is no tangent of that gradient.

A locus is the path of a point $P(x,y)$ that obeys a rule. Write the rule using coordinates, then simplify. Distance rules. $PA=k\,PB$, or $PS=e\times$(distance to a line). Square both sides to remove the root. For $PS=\frac12\times$ distance from $x=12$ with $S(3,0)$: $4\left[(x-3)^2+y^2\right]=(x-12)^2\Rightarrow3x^2+4y^2=108\Rightarrow\frac{x^2}{36}+\frac{y^2}{27}=1,$ an ellipse (this is the focus and directrix property with eccentricity $\frac12$). Parametric rules. Eliminate the parameter, using $\cos^2\theta+\sin^2\theta=1$ or $\sec^2\theta-\tan^2\theta=1$. Check. Identify the shape (circle, ellipse, line) and test one point that should satisfy the rule.