FP3.2 - Further coordinate systems

A standard ellipse or hyperbola is centred at the origin with its transverse or major axis on the $x$-axis. Assume $a>0$ and $b>0$. The signs in the Cartesian equation identify the curve: an ellipse has a sum equal to $1$, while a hyperbola has a difference equal to $1$. A parametric pair describes points on the curve using one parameter and is often the cleanest form for differentiation or moving-point problems.

Verify a parametrisation by substitution. For the ellipse,$$\frac{(a\cos t)^2}{a^2}+\frac{(b\sin t)^2}{b^2}=\cos^2t+\sin^2t=1.$$For the trigonometric hyperbola parametrisation, the corresponding identity is $\sec^2t-\tan^2t=1$; for the hyperbolic parametrisation it is $\cosh^2t-\sinh^2t=1$. This check also explains why the same parameter must appear in both coordinates.

To remove a parameter, first isolate functions of it and then use the matching identity. If $x=4\cos t$ and $y=3\sin t$, then $\cos t=x/4$ and $\sin t=y/3$, so$$\frac{x^2}{16}+\frac{y^2}{9}=1.$$The coordinate bounds $|x|\le4$ and $|y|\le3$ follow from the parametrisation as well as from the ellipse.

A Cartesian equation is usually better for intersections. Substitute the equation of the line into the conic to obtain a quadratic in one coordinate. Its two roots represent the two intersection coordinates; their sum and product can locate a midpoint or build a symmetric expression without solving for each point separately. Return to the line equation for the other coordinate, and check that every resulting point lies on both curves.

If a conic is translated, replace $x$ and $y$ by displacements from its centre. For example,$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$has parametrisation $x=h+a\cos t$, $y=k+b\sin t$. Complete squares before reading a centre or semi-axis from a non-standard Cartesian equation.

For a fixed focus $S$ and directrix $\ell$, a conic is the locus of points $P$ satisfying$$PS=e\,d(P,\ell),$$where $e>0$ is the eccentricity and $d(P,\ell)$ is the perpendicular distance to the line. An ellipse has $0<e<1$ and a hyperbola has $e>1$. For the standard horizontal conics, the foci and directrices occur in symmetric pairs on the $x$-axis.

The focus is farther from the centre than the corresponding vertex for a hyperbola because $e>1$, but nearer for an ellipse because $e<1$. Thus the quantities $ae$ and $a/e$ switch their relative sizes: for an ellipse $ae<a<a/e$, whereas for a hyperbola $a/e<a<ae$. This is a useful geometric check on calculated coordinates and directrices.

The ellipse relation can be recovered from the locus definition. Use the right focus $S=(ae,0)$ and right directrix $x=a/e$. For $P=(x,y)$ on the appropriate side of that directrix,$$(x-ae)^2+y^2=e^2\left(x-\frac ae\right)^2.$$Expansion cancels the linear terms and gives$$(1-e^2)x^2+y^2=a^2(1-e^2).$$Dividing through and comparing with $x^2/a^2+y^2/b^2=1$ yields $b^2=a^2(1-e^2)$. The same distance principle, with $e>1$, gives the hyperbola relation.

Given any two compatible focus-directrix quantities, solve systematically. If a hyperbola has focus distance $ae=10$ and directrix distance $a/e=5/2$, then $a^2=(ae)(a/e)=25$, so $a=5$, $e=2$, and$$b^2=a^2(e^2-1)=75.$$Its standard equation is therefore $x^2/25-y^2/75=1$. The positive square root is used because eccentricity and semi-axis length are positive.

These displayed formulae assume a horizontal major or transverse axis. For a vertical conic, the roles of $x$ and $y$ swap: the foci lie at $(0,\pm ae)$ and the directrices are $y=\pm a/e$. Identify the axis from the equation before applying a memorised coordinate pattern.

A tangent has the curve's instantaneous gradient at the point of contact; a normal is perpendicular to it. There are two complementary routes. Use differentiation when the contact point is known or parametrised. Use a repeated-root condition when a line $y=mx+c$ is required to touch a conic at exactly one point.

For the ellipse $x^2/a^2+y^2/b^2=1$, implicit differentiation gives$$\frac{dy}{dx}=-\frac{b^2x}{a^2y}.$$At $(a\cos t,b\sin t)$ this is $-b\cos t/(a\sin t)$. Substitution in point-gradient form and simplification produces the tangent in the table. The normal gradient is the negative reciprocal, $a\sin t/(b\cos t)$, provided neither relevant gradient is vertical.

Parametric differentiation gives the same result without first eliminating $t$:$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$$For $x=a\sec t$, $y=b\tan t$, this gives $dy/dx=b\sec^2t/(a\sec t\tan t)$. Keep the contact point attached to the same parameter value when writing the line through it.

These conditions for $y=mx+c$ come from substituting the line into the Cartesian conic and setting the resulting quadratic discriminant to zero. For example, tangents of slope $3/4$ to $x^2/16+y^2/9=1$ satisfy$$c^2=16\left(\frac34\right)^2+9=18,$$so the two parallel tangents are $y=\tfrac34x\pm3\sqrt2$. A line through a specified external point supplies a second equation relating $m$ and $c$.

After finding a tangent, recover its contact point by solving the line and conic simultaneously. Because the intersection is a repeated root, the quadratic should have one repeated coordinate; substitution gives the other coordinate. This is also a check that the proposed line really touches rather than cuts the curve.

A locus records every position a moving point can occupy. In a coordinate problem, first express the moving point as $(x(t),y(t))$ from the geometry; then eliminate $t$ to obtain a Cartesian relation. The relation is not the whole answer until restrictions inherited from the parameter and construction have been checked.

Suppose a moving midpoint has coordinates$$x=2+3\cos t,\qquad y=-1+2\sin t.$$Then $\cos t=(x-2)/3$ and $\sin t=(y+1)/2$. Using $\cos^2t+\sin^2t=1$ gives the locus$$\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1.$$If $t$ ranges through a full $2\pi$ interval, the whole ellipse is traced; a smaller interval would select only part of it.

When coordinates are rational functions of a parameter, avoid solving unnecessarily complicated equations. If$$x=\frac{6u}{u^2+1},\qquad y=\frac{3(1-u^2)}{u^2+1},$$then direct calculation gives $x^2/9+y^2/9=1$. The missing or repeated points must still be checked from the original expressions; algebraic elimination alone does not record how the curve is traced.

Points created from intersections, midpoints or perpendicular lines should be built in layers. Find the defining lines or intersection coordinates first; use root sums for a midpoint when two intersections are roots of one quadratic; then eliminate the moving parameter. This keeps each geometric condition visible and prevents an unexplained leap to the final equation.

After elimination, complete squares and normalise to recognise the curve. For example,$$4x^2+9y^2+18y=27$$becomes$$\frac{x^2}{9}+\frac{(y+1)^2}{4}=1.$$A correct locus equation should be tested by substituting the original parametric coordinates back into it identically.