Re: Re: Needed Grid lines in Implicit Plot

```On 13 Dec 2004, at 18:24, Bob Walker wrote:

> Remove ", {y, -4 , 4 }" iterator and it will work as you wish.
> ImplicitPlot accepts only one iterator.  You can inspect the
> ImplicitPlot signature by:
> ?ImplicitPlot
>
> I don't know why there is no error indication.
>
> Take care,
> Bob Walker
>
> Narasimham wrote:
>
>> << Graphics`ImplicitPlot`
>> ImplicitPlot[3 x + 5 y == 1, {x, -4 , 4}, {y, -4 , 4 },
>> GridLines ->{Range[-4, 4, 1], Range[-4, 4, 1]},AspectRatio ->
>> Automatic];
>>
>>

This is quite simply false.

There are two forms of Implicitplot. they are both described int he

ImplicitPlot[eqn, {x, a, b}] draws a graph of the set of points that
satisfy \
the equation eqn.  The variable x is associated with the horizontal
axis and \
ranges from a to b.  The remaining variable in the equation is
associated \
with the vertical axis. ImplicitPlot[eqn, {x, a, x1, x2, ..., b}]
allows the \
user to specify values of x where special care must be exercised. \
ImplicitPlot[{eqn1, eqn2, ...}, {x, a, b}] allows more than one
equation to \
be plotted, with PlotStyles set as in the Plot function.
ImplicitPlot[eqn, \
{x, a, b}, {y, a, b}] uses a contour plot method of generating the
plot.  \
This form does not allow specification of intermediate points

The form of ImplicitPLot with one iterator only works on equations that
can be "solved" algebraically using Solve. it produces a Graphics
object, which is why it accepts the GridLines option.

The other form of ImplicitPlot works much more generally, but it
produces a CountourGraphics object, which needs to be converted to a
Graphics object before using the GridLines option.

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/

```

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