FP1.4 - Coordinate systems

Learning objectives

—FP1.4.1 - Cartesian equations for the parabolaCartesian equations for the parabola Students should be familiar with the equations: and rectangular hyperbola. c y2 = 4ax or x = at2, y = 2at and xy = c2 or x = ct, y =. t.—FP1.4.2 - Parametric equations for conicsIdea of parametric equation for The idea of (at2, 2at) as a general point on the parabola is parabola and rectangular hyperbola. all that is required.—FP1.4.3 - Focus-directrix property of the parabolaThe focus-directrix property of the Concept of focus and directrix and parabola as locus of parabola. points equidistant from focus and directrix.—FP1.4.4 - Tangents and normals to conicsTangents and normals to these 1 1 c2 Differentiation of y = 2a2x2, y =. curves. x Parametric differentiation is not required.

The FP1 standard conics are a parabola with its vertex at the origin and axis along the $x$ -axis, and a rectangular hyperbola whose asymptotes are the coordinate axes. Their constants control scale, not a translation of the centre or vertex.

Curve	Cartesian equation	Key features
parabola	$y^2=4ax$	vertex $(0,0)$ ; axis $y=0$ ; for $a>0$ it opens to the right
rectangular hyperbola	$xy=c^2$ , or $y=c^2/x$	asymptotes $x=0$ and $y=0$ ; branches lie where $x$ and $y$ have the same sign

For $y^2=12x$ , comparison with $y^2=4ax$ gives $a=3$ . For $xy=25$ , comparison with $xy=c^2$ gives $c=5$ or $c=-5$ ; the curve depends on $c^2$ , so the conventional positive scale is $|c|=5$ .

Do not confuse $4a$ with $a$ , or $c^2$ with $c$ . A rectangular hyperbola never meets either coordinate axis because $xy=c^2
e0$. These standard equations are not the general translated or rotated forms of every parabola or hyperbola.

A parameter replaces the two coordinates of a point by one variable. Substitution verifies that the parameterised point lies on the conic and lets the same algebra describe every allowed point.

Curve	General point	Verification
$y^2=4ax$	$(at^2,2at)$ , $t\in\mathbb R$	$(2at)^2=4a(at^2)$
$xy=c^2$	$(ct,c/t)$ , $t\ne0$	$(ct)(c/t)=c^2$

On $y^2=12x$ , $a=3$ . At $t=2$ , the point is $(3\cdot2^2,2\cdot3\cdot2)=(12,12)$ . Conversely, a point with $y=12$ has $t=y/(2a)=2$ . On $xy=25$ , taking $c=5$ and $t=-2$ gives $(-10,-5/2)$ , whose coordinate product is $25$ .

The same letter $t$ labels a point; it is not a coordinate or a fixed curve constant. The hyperbola excludes $t=0$ . For this unit, understanding and using the general points is required, but parametric differentiation is not.

A parabola is the locus of points whose distance from a fixed point, the focus, equals their perpendicular distance from a fixed line, the directrix.

$y^2=4ax:\qquad \text{focus }(a,0),\qquad \text{directrix }x=-a$

For $P=(x,y)$ on $y^2=4ax$ with $a>0$ , $PF^2=(x-a)^2+y^2=(x-a)^2+4ax=(x+a)^2.$ Since the perpendicular distance from $P$ to $x=-a$ is $x+a$ , the two distances are equal. This also places the vertex midway between focus and directrix at $(0,0)$ .

For $y^2=20x$ , $a=5$ , so the focus is $(5,0)$ and the directrix is $x=-5$ . The point $(5,10)$ lies on the curve. Its distance from the focus is $10$ , and its perpendicular distance from the directrix is also $5-(-5)=10$ .

Distance to a line means the shortest, perpendicular distance—not distance to an arbitrary point on the line. For $x=-a$ it is the horizontal distance. Do not place the directrix at $x=a$ or the focus at $(4a,0)$ .

Find a tangent gradient by differentiating the Cartesian equation, then use the negative reciprocal for the normal gradient. A parameter may identify the point, but parametric differentiation is not required.

Curve and point	Tangent gradient	Tangent	Normal
$y^2=4ax$ , $P=(at^2,2at)$	$1/t$	$ty=x+at^2$	$y=-tx+2at+at^3$
$xy=c^2$ , $P=(ct,c/t)$	$-1/t^2$	$x+t^2y=2ct$	$t^3x-ty=c(t^4-1)$

For the parabola, $2y\,dy/dx=4a$ , so $dy/dx=2a/y=1/t$ at $P$ . For the hyperbola, $y=c^2x^{-1}$ gives $\frac{dy}{dx}=-\frac{c^2}{x^2}=-\frac1{t^2}.$ Insert each point and gradient into $y-y_0=m(x-x_0)$ , then simplify.

For $y^2=8x$ , $a=2$ . At parameter $t=3$ , $P=(18,12)$ . The tangent is $3y=x+18,$ and the normal is $y=-3x+66.$ Their gradients $1/3$ and $-3$ multiply to $-1$ , checking perpendicularity.

Do not differentiate $x=at^2$ and $y=2at$ with respect to $t$ in this unit. The gradient formulas containing $1/t$ exclude the parabola's vertex $t=0$ ; there the tangent is the vertical line $x=0$ and the normal is $y=0$ .

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—FP1.5 - Matrix algebraRead

FP1.4 - Coordinate systems

Learning objectives

Recognise the standard conics

Use a parameter to name a point on a conic

Define a parabola by focus and directrix

Construct tangents and normals to the conics

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