Parametric Equations of the Parabola

Equations can also be written in parametric form. In this form, x and y are both written in terms of a third variable called a parameter.

An example of a parametric equation is x=2t, y=t-2.

Try to write y=3x+1 in a parametric form.

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Given parameter p:

Let x = p

Then y = 3x + 1

So y = 3p + 1

 

There are many ways to write parametric equations, as you can use different parameters other than p. We can also change parametric equations back into Cartesian form.

For example,

Find the Cartesian equation of x = 3t + 1y = 2t - 3.

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We use the process for solving simultaneous equations to eliminate the parameter.

x=3t+1\qquad (1)\\ y=2t-3\qquad (2)\\ \\ From\quad (1):\\ x-1=3t\\ \\ t=\cfrac { x-1 }{ 3 } \\ \\ Substitute\quad in\quad (2):\\ \\ y=2t-3\\ \\ y=2\left( \cfrac { x-1 }{ 3 } \right) -3\\ \\ 3y=2(x-1)-9\\ 3y=2x-2-9\\ 3y=2x-11\\ 2x-3y-11=0

 

 

Parametric Representation of a Curve

 

For the equation of a parabola {x}^{2}=4ay, the parametric equation can be written as:

x=2at \\ y=a{t}^{2}

where t is a parameter.

Therefore, the other types of parabola can be written as below.

 

Parabola {x}^{2}=-4ay can be written as:

x=2at \\ y=-a{t}^{2}

 

Parabola {y}^{2}=4ax can be written as:

x=a{t}^{2}\\ y=2at

 

Parabola {y}^{2}=-4ax can be written as:

x=-a{t}^{2}\\ y=2at

 

 

Examples

 

1. Given the parabola x=4t and y=2{t}^{2}, find:

a) its Cartesian equation

b) the points on the parabola when t=\pm 2

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a)

x=4t\\ \\ t=\cfrac { x }{ 4 } \\ \\ Subs\quad into\quad y=2{ t }^{ 2 }\\ \\ y=2{ \left( \cfrac { x }{ 4 } \right) }^{ 2 }\\ \\ y=\cfrac { 2{ x }^{ 2 } }{ 16 } \\ \\ y=\cfrac { { x }^{ 2 } }{ 8 } \\ \\ 8y={ x }^{ 2 }

b)

When\quad t=2\\ \\ x=4(2)=8\\ y=2{ \left( 2 \right) }^{ 2 }=8\\ \\ When\quad t=-2\\ \\ x=4(-2)=-8\\ y=2{ \left( -2 \right) }^{ 2 }=8\\ \\ \therefore \quad Points\quad are\quad (8,8)\quad and\quad (-8,8)

 

2. Find the coordinates of the focus and the equation of the directrix of the parabola x=-12t, \ y=-6{t}^{2}.

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x=-12t\qquad (1)\\ y=-6{ t }^{ 2 }\qquad (2)\\ \\ From\quad (1):\\ x=-12t\\ \\ t=-\cfrac { x }{ 12 } \\ \\ Subs\quad in\quad (2)\\ y=-6{ t }^{ 2 }\\ \\ y=-6{ \left( -\cfrac { x }{ 12 } \right) }^{ 2 }\\ \\ y=-6{ \left( \cfrac { { x }^{ 2 } }{ 144 } \right) }\\ \\ y=-\cfrac { { x }^{ 2 } }{ 24 } \\ \\ { x }^{ 2 }=-24y

The form of this curve is concave downwards parabola with vertex at the origin.

4a=24 \\ a=6

So focal length is 6 units, and the directrix is y = 6.

Chords, Tangents and Normals

 

A number of parabola parametric properties are below:

 

If P(2ap,a{ p }^{ 2 }) and Q(2aq,a{ q }^{ 2 }) are any two points on the parabola {x}^{2}=4ay, then the chord PQ has gradient \cfrac { p+q }{ 2 } and equation y-\cfrac { 1 }{ 2 } \left( p+q \right) x+apq=0.

 

If PQ is a focal chord, then pq = -1.

 

The tangent to the parabola {x}^{2}=4ay at the point P(2ap,a{ p }^{ 2 }) has gradient p and equation given by y-px+a{ p }^{ 2 }=0.

 

The tangents to the parabola {x}^{2}=4ay at points P(2ap,a{ p }^{ 2 }) and Q(2aq,a{ q }^{ 2 }) intersect at the point [a(p+q), apq].

 

The normal to the curve {x}^{2}=4ay at point P(2ap,a{ p }^{ 2 }) has gradient - \cfrac{1}{p} and equation given by x+py=a{ p }^{ 3 }+2ap

 

The normals to the parabola {x}^{2}=4ay at P(2ap,a{ p }^{ 2 }) and Q(2aq,a{ q }^{ 2 }) intersect at [-apq(p+q),a({ p }^{ 2 }+pq+{ q }^{ 2 }+2)]

 

Examples

 

1. Find the equation of the chord joining points where t=3 and t=-2 on the parabola x=2at,\quad y=a{ t }^{ 2 }.

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When\quad t=3\\ x=2a(3)\qquad \qquad y=a{ \left( 3 \right) }^{ 2 }\\ x=6a\qquad \qquad \quad y=9a\\ \\ When\quad t=-2\\ x=2a(-2)\qquad \qquad y=a{ \left( -2 \right) }^{ 2 }\\ x=-4a\qquad \qquad \quad y=4a\\ \\ \therefore \quad Points\quad are\quad (6a,\quad 9a)\quad and\quad (-4a,\quad 4a)\\ \\ \\ Gradient:\\ \\ m=\cfrac { { y }_{ 2 }-{ y }_{ 2 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \\ m=\cfrac { 4a-9a }{ -4a-6a } \\ \\ m=\cfrac { -5a }{ -10a } \\ \\ m=\cfrac { 1 }{ 2 } \\ \\ Using\quad point-gradient:\\ y-{ y }_{ 1 }=m(x-{ x }_{ 1 })\\ \\ y-4a=\cfrac { 1 }{ 2 } (x+4a)\\ \\ 2y-8a=x+4a\\ 0=x-2y+12a

 

2. Find the equation of the tangent to the parabola {x}^{2}=8y at the point (4t,2{t}^{2}).

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{ x }^{ 2 }=8y\\ \\ \therefore \quad y=\cfrac { { x }^{ 2 } }{ 8 } \\ \\ \cfrac { dy }{ dx } =\cfrac { 2x }{ 8 } =\cfrac { x }{ 4 } \\ \\ At\quad \left( 4t,2{ t }^{ 2 } \right) \\ \\ \cfrac { dy }{ dx } =\cfrac { 4t }{ 4 } =t\\ \\ y-{ y }_{ 1 }=m\left( x-{ x }_{ 1 } \right) \\ \\ y-2{ t }^{ 2 }=t\left( x-4t \right) \\ y-2{ t }^{ 2 }=tx-4{ t }^{ 2 }\\ 0=tx-y-2{ t }^{ 2 }

 

Another approach to the equations of tangent, normal and chord of a parabola is derived from points in Cartesian form.

 

If point A\left( { x }_{ 1 },{ y }_{ 1 } \right) lies on the parabola { x }^{ 2 }=4ay, then the equation of the tangent at A is given by:

x{ x }_{ 1 }=2a(y+{ y }_{ 1 })

 

If point A\left( { x }_{ 1 },{ y }_{ 1 } \right) lies on the parabola { x }^{ 2 }=4ay, then the equation of the normal at A is given by:

y-{ y }_{ 1 }=-\cfrac { 2a }{ { x }_{ 1 } } \left( x-{ x }_{ 1 } \right)

 

The equation of the chord of contact XY of tangents drawn from external point \left( { x }_{ 1 },{ y }_{ 1 } \right) to the parabola { x }^{ 2 }=4ay is given by:

x{ x }_{ 1 }=2a(y+{ y }_{ 1 })

 

Example

 

Tangents are drawn from the point \left( \cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 2 } \right) to the points P and Q on the parabola {x}^{2}=4y. Find the equation of the chord of contact PQ and the coordinates of P and Q.

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{ x }^{ 2 }=4y\\ 4a=4\\ a=1\\ \\ PQ\quad \rightarrow \quad x{ x }_{ 1 }=2a(y+{ y }_{ 1 })\quad at\quad \left( \cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 2 } \right) \\ \\ \cfrac { 1 }{ 2 } x=2\left( y-\cfrac { 1 }{ 2 } \right) \\ \\ \cfrac { 1 }{ 2 } x=2y-1\\ \\ x=4y-2\\ x-4y+2=0\quad is\quad the\quad chord\quad of\quad contact\\ \\ Find\quad P\quad and\quad Q:\\ \\ { x }^{ 2 }=4y\qquad \qquad (1)\\ x-4y+2=0\qquad (2)\\ \\ From\quad (2):\qquad x+2=4y\\ \\ Subs\quad into\quad (1):\\ { x }^{ 2 }=x+2\\ { x }^{ 2 }-x-2=0\\ (x-2)(x+1)=0\\ x=2,-1\\ \\ Substitute\quad x=2\quad into\quad (3)\\ \\ 2-4y+2=0\\ 4y=4 \\ y=1

Substitute\quad x=-1\quad into\quad (3)\\ \\ -1-4y+2=0\\ 4y=1\\ y=\cfrac { 1 }{ 4 } \\ \\ So\quad P\quad and\quad Q\quad are\quad points:\\ \\ \left( 2,1 \right) \quad and\quad \left( -1,\cfrac { 1 }{ 4 } \right)

 

More Properties of the Parabola

 

The tangents at the end of a focal chord intersect at right angles on the directrix.

For example,

Points P\left( 2,-\cfrac { 1 }{ 2 } \right) and Q\left( -8,-8 \right) lie on the parabola { x }^{ 2 }=-8y.

a) Find the equation of line PQ.

b) Show that PQ is a focal chord

c) Prove that the tangents at P and Q intersect at right angles on the directrix.

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a) Equation of PQ:

\cfrac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } =\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \\ \cfrac { y+8 }{ x+8 } =\cfrac { -\frac { 1 }{ 2 } +8 }{ 2+8 } \\ \\ \cfrac { y+8 }{ x+8 } =\cfrac { 3 }{ 4 } \\ \\ 4y+32=3x+24\\ 3x-4y-8=0\qquad (1)

b)

{ x }^{ 2 }=-8y\\ 4a=8\\ a=2\\ Focus=(0,-2)\\ Substitute\quad (0,-2)\quad into\quad (1)\\ RHS=3(0)-4(-2)-8\\ \qquad =0\\ \qquad =LHS\\ \therefore \quad PQ\quad is\quad a\quad focal\quad chord

c)

{ x }^{ 2 }=-8y\\ \\ y=-\cfrac { { x }^{ 2 } }{ 8 } \\ \\ \cfrac { dy }{ dx } =-\cfrac { x }{ 4 } \\ \\ At\quad P,\quad x=2,\quad y=-\cfrac { 1 }{ 2 } ,\quad so,\\ \\ \cfrac { dy }{ dx } =-\cfrac { 2 }{ 4 } =-\cfrac { 1 }{ 2 } \\ \\ \\ Using\quad point-gradient\quad form:\\ y-{ y }_{ 1 }=m(x-{ x }_{ 1 })\\ \\ y+\cfrac { 1 }{ 2 } =-\cfrac { 1 }{ 2 } (x-2)\\ \\ 2y+1=-x+2\\ x+2y-1=0\qquad \qquad (1)\\ \\ At\quad Q,\quad x=-8,\quad y=-8,\quad so,\\ \\ \cfrac { dy }{ dx } =-\cfrac { (-8) }{ 4 } =2\\ \\ y+8=2(x+8)\\ y+8=2x+16\\ 2x-y+8=0\qquad \qquad (2)\\ \\ Hence,\quad check\quad { m }_{ 1 }{ m }_{ 2 }=-1:\\ \\ -\cfrac { 1 }{ 2 } \times 2=-1\\ \\ \therefore \quad The\quad tangents\quad are\quad perpendicular.\\ \\ \\ Finding\quad the\quad point\quad of\quad intersection:\\ \\ \begin{matrix} x & + & 2y & - & 1 & = & 0 & \qquad & (1) & \times 2 \\ 2x & - & y & + & 8 & = & 0 & \qquad & (2) & \quad \end{matrix}\\ \\ \begin{matrix} 2x & + & 4y & - & 2 & = & 0 & \qquad & (3) \\ 2x & - & y & + & 8 & = & 0 & \qquad & \quad \end{matrix}\\ \\ (2)-(3):\\ -5y+10=0\\ 5y=10\\ y=2\\ \\ Substitute\quad in\quad (1)\\ x+4-1=0\\ x=-3\\ \\ \therefore \quad Point\quad of\quad intersection\quad is\quad (-3,2)

The directrix has equation y = a, that is y = 2.

The point (-3, 2) lies on the line y = 2, which is the directrix.

 

Another property of the parabola is that any point P on a parabola is equally inclined to the axis of the parabola and the focal chord through P.