Question 767517
<pre>
This conic appears at first to be an ellipse since there is
no xy term, an x² and a y² term with different coefficients
with the same sign when on the same side of the equation.  
So we try to get it in standard form.  Either
in the form {{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} or {{{(x-h)^2/b^2+(y-k)^2/a^2=1}}} 

   4x² + 5y² - 8x + 20y = -24

   4x² - 8x + 5y² + 20y = -24

4(x² - 2x) + 5(y² + 4y) = -24

To complete the square in the first parentheses,
Multiply the coefficient of x, which is -2, by 1/2
getting -1.  Then square -1, which gives +1. So we
add +1 to the right of the first parentheses, and
since there is a 4 multiplied by the first parentheses
we add 4·1 or 4 to the right side of the equation:

4(x² - 2x + 1) + 5(y² + 4y) = -24 + 4

To complete the square in the second parentheses,
Multiply the coefficient of y, which is 4, by 1/2
getting 2.  Then square 2, which gives 4. So we
add 4 to the right of the second parentheses, and
since there is a 5 multiplied by the second parentheses
we add 5·4 or 20 to the right side of the equation:

4(x² - 2x + 1) + 5(y² + 4y + 4) = -24 + 4 + 20

Factor the first parentheses as (x - 1)(x - 1) or (x - 1)²
Factor the second parentheses as (x - 2)(x - 2) or (x - 2)²
Combine the numbers on the right, getting 0.

4(x - 1)² + 5(y + 2) = 0

Oh oh.  This is an unusual situation because the right
side came out to be 0.  So we cannot get 1 on the right
side as we could if it were some other number.

This means that the graph is of an ellipse that is
degenerated into a single point.  The graph of the
equation is simply a single point (1,-2)

The graph is simply this single point (1,-2):

{{{drawing(400,400,-3,3,-3,3, graph(400,400,-3,3,-3,3),
circle(1,-2,.02),circle(1,-2,.03),circle(1,-2,.04),circle(1,-2,.05) )}}}

This equation is the equation of a single point.  A single point
IS a conic section, because the vertex of a cone is a single point.
A plane intersecting a cone at its vertex is a single point. 

Edwin</pre>