Graphing conic sections, summary:

An equation of the form

alpha x^2 + beta x + gamma y^2 + delta y + lambda = 0

is quadratic in both x and y.  Depending on the values of alpha and gamma (the coefficients of x^2 and y^2) this equation can yield a straight line, a parabola, an ellipse or a hyperbola.

It is not difficult to find and graph the curve after completing the square on x if alpha is nonzero, and on y if beta is nonzero.

 

y = x^2 is the basic parabola, vertex at (0, 0), going through points (-1, 1) and (1, 1), axis of symmetry the y axis.

y = A x^2 is the basic parabola stretched vertically by factor A, vertex at (0, 0), going through points (-1, A) and (1, A).

y - k = A (x - h)^2 is the parabola y = A x^2 shifted horizontally h units and vertically k units.  Its vertex is at (h, k) and it passes through the points (h + 1, k + A) and (h - 1, k + A).

y = a x^2 + b x + c can be put into the preceding form by completing the square on x, and moving any remaining constant to the y side of the equation.

The zeros of y = a x^2 + b x + c occur when a x^2 + b x + c = 0; by the quadratic formula they occur at x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a).  The line x = - b / (2 a) is halfway between the zeros, if they exist; and whether the zeros exist or not this line is the axis of symmetry.  The vertex lies at x coordinate -b / ( 2 a).  The y coordinate of the vertex is found by plugging the x coordinate into the equation.  If the y coordinate of the vertex is y_0, then at x = -b / (2 a) + 1 and at x = -b / (2 a) - 1 the y coordinate is y_0 + a, where a is the coefficient of x^2.

x = y^2 is the basic parabola with vertex at (0, 0), going through points (1, 1) and (1, -1), axis of symmetry the x axis.

Its general form is x - h = A ( y - k )^2, with vertex at (h, k), and it passes through the points (h + A, k - 1) and (h + A, k + 1).

These parabolas occur as a result of the form alpha x^2 + beta x + gamma y^2 + delta y + lambda = 0 when one, but not both, of alpha and gamma (the coefficients of x^2 and y^2) is zero.

 

This can be generalized to the x-z or y-z planes in 3-dimensional space, where we can get parabolas like z - k = A (x - h)^2, or x - h = A ( z - k)^2, or z - k = A ( y - h)^2, or y - h = A ( z - k )^2.

Similar generalizations apply to ellipses and hyperbolas.

 

x^2 + y^2 = 1 is the basic circle of radius 1 centered at the origin.

x^2 / r^2 + y^2 / r^2 = 1 would be a circle of radius r centered at the origin.

x^2 / a^2 + y^2 / b^2 = 1 would be an ellipse centered at the origin, with semiaxes a in the x direction and b in the y direction.

(x-h)^2 / a^2 + (y- k)^2 / b^2 = 1 is an ellipse centered at the (h, k), with semiaxes a in the x direction and b in the y direction.

A convenient way to visualize and sketch the ellipse is to draw the rectangle bounded by the lines x = h - a, x = h + a, y = k -b, y = k + b.  This rectangle is centered at (h, k) and has dimensions 2 a by 2 b, with 2 a in the x direction and 2 b in the y direction.  The ellipse is tangent to each side of this rectangle, with points of tangency being (h, k + b), (h, k - b), (h + b, k) and (h - b, k).

In general an ellipse results from the form alpha x^2 + beta x + gamma y^2 + delta y + lambda = 0 when alpha and gamma, the coefficients of x^2 and y^2, are both nonzero and both of the same sign.

 

x^2 - y^2 = 1 and -x^2 + y^2 = 1 are the basic hyperbolas.  They can be sketched by first sketching the square bounded by the lines x = -1, x = 1, y = -1 and y = 1, then sketching the diagonals and extending them to infinity in both directions.  Figure out on which axis the vertices of the hyperbola lies.  A vertex occurs when x = 0 or y = 0.  For x^2 - y^2 = 1, x can't be zero since -y^2 is negative and 1 is positive.  y can be zero, which gives you x = +- 1.  So the vertices are (-1, 0) and (1, 0).  For -x^2 + y^2 = 1, y can't be zero so the vertices are at (0, 1) and (0, -1).  The hyperbola will have two unconnected parts, one part passing through each vertex and being asymptotic to the extended diagonals.

x^2 / a^2 - y^2 / b^2 = 1 and -x^2 / a^2 + y^2 / b^2 = 1 are sketched within boxes bounded by x = a, x = -a, y = b and y = -b.  The diagonals are again extended.  The vertices of x^2 / a^2 - y^2 / b^2 = 1 are at (-a, 0) and (a, 0).  The vertices of -x^2 / a^2 + y^2 / b^2 = 1 are at (0, -b) and (0, b).

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 and -(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1 are sketched with the aid of the rectangle bounded by the lines x = h - a, x = h + a, y = k -b, y = k + b (the same rectangle used for sketching the general ellipse).

In general an ellipse results from the form alpha x^2 + beta x + gamma y^2 + delta y + lambda = 0 when alpha and gamma, the coefficients of x^2 and y^2, are both nonzero and of opposite signs.

Application to quadric surfaces

Consider first the surface z = 4 - x^2 - y^2.

What is the x-z trace of this surface (the x-z trace is the graph of the surface, confined to the x-z plane; we confine the graph to the x-z plane by letting y = 0)?  What are the coordinates of the vertex, the x = 1 and the x = -1 points of this graph?

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Can you sketch a graph of the x-z trace on an x-z coordinate system?  Begin by sketching the vertex and the x = 1 and x = -1 points of the parabola.

Can you sketch this graph as it would appear within an x-y-z coordinate system?  Begin by sketching the vertex and the x = 1 and x = -1 points of the parabola.  With these points to guide you it shouldn't be difficult to make a reasonable sketch.

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What is the y-z trace of this surface (the y-z trace is the graph of the surface, confined to the y-z plane; we confine the graph to the y-z plane by letting y = 0)?  What are the coordinates of the vertex, the y = 1 and the y = -1 points of this graph?

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Can you sketch a graph of the y-z trace on an y-z coordinate system?  Begin by sketching the vertex and the y = 1 and y = -1 points of the parabola.

Can you sketch this graph as it would appear within an x-y-z coordinate system?  Begin by sketching the vertex and the y = 1 and y = -1 points of the parabola.  With these points to guide you it shouldn't be difficult to make a reasonable sketch.

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What is the x-y trace of this surface?

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Can you sketch the x-y trace in an x-y coordinate system?  Begin by finding and sketching the points where x = 0, and the points where y = 0.

Can you sketch the x-y trace in an x-y-z coordinate system?  Begin by finding and sketching the points where x = 0, and the points where y = 0.  With these points to guide you it shouldn't be difficult to make a reasonable sketch.

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Let's now consider the plane z = 3.  This plane is parallel to and lies 3 units above the x-y plane.  It includes the points (0, 0, 3), (1, 0, 3), (-1, 0, 3), (0, -1, 3) and (0, 1, 3).  Can you sketch these points in and x-y-z coordinate system?  The 'auxiliary' x axis within this plane is a straight line through the points (0, 0, 3), (1, 0, 3), (-1, 0, 3).  Sketch it.  Then sketch the 'auxiliary' y axis for this plane.

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If z = 3, what is the equation z = 4 - x^2 - y^2?  What conic section does this define in x and y?  Sketch this conic section in an x-y plane.  Then sketch it again in x-y-z space, using the 'auxiliary' x and y axes you have constructed above.

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Now consider the plane x = 1.  Sketch this plane on an x-y-z coordinate system.  Begin by sketching the point (1, 0, 0), which will serve as the origin of an auxiliary y-z plane.  Then sketch an auxiliary y axis, and an auxiliary z axis.

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Substitute x = 1 into the equation z = 4 - x^2 - y^2.  You will get an equation in y and z, and the equation gives you a conic section in y and z.  Sketch this conic section in a y-z plane, being sure to first sketch at least 3 basic points as a guide if you have a parabola (e.g., the vertex and the y = 1 and y = -1 points), or the four corners of the 'box' used to sketch an ellipse or a hyperbola.  Then make the same sketch in your x-y-z coordinate system, using your auxiliary axes (and sketching the basic points, or the 'box') as a guide.

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Next follow the analogous procedure for the plane y = 1.

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Go back to your sketches for the plane x = 1.  Look at the two-dimensional sketch you made for the y-z plane.  Sketch in the unit vectors `j and `k, each vector originating from the point (0, 0). 

Find the y = 1 point on your graph, and find the slope of your graph at that point.  Sketch a vector with `j component 1, in the direction of the tangent line, originating at the y = 1 point of your curve.  What is the `k component of your vector?

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Now go to your sketch in the x-y-z system, and sketch the same vector.

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Now go to your sketch for the plane y = 1.  In your two-dimensional sketch, sketch in the `i and `k vectors, each originating from (0, 0).  Then find the slope of the graph at the x = 1 point, and sketch the vector with `i component 1, in the direction of the tangent line, with the vector originating at the x = 1 point of your curve.  Then sketch the same vector in your x-y-z sketch of this curve.

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You have sketched two vectors in x-y-z coordinate systems.  What is the cross product of these vectors, and what is the significance of this cross product?

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What is the partial derivative z_x of the the function z(x, y) = 4 - x^2 - y^2?  What is z_y?

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What is the value of each of these partial derivatives, evaluated at the point (x, y) = (1, 1)?

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What do these partial derivatives have to do with your graphs?

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What do these partial derivatives have to do with the vectors you have sketched?

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