Conic sections

Classification takes two questions, asked in order:

1. How many squared variables? One squared variable (the other appearing linearly) means a parabola, like $y = a(x-h)^2 + k$ or $x = a(y-k)^2 + h$.
2. Two squared: what sign sits between them? Added with equal coefficients: a circle. Added with unequal positive coefficients: an ellipse. Subtracted: a hyperbola, no matter what the coefficients are.

The sign between the squared terms is the load-bearing detail. $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and $\frac{x^2}{9} - \frac{y^2}{4} = 1$ differ by one character and are entirely different animals: a closed oval versus two open branches with asymptotes. "Both variables are squared, so it is an ellipse" is the pattern-match that dies here.

The circle's standard form is $(x - h)^2 + (y - k)^2 = r^2$, center (h, k), radius r. Two misreads account for most lost points:

The signs flip. $(x - 2)^2$ places the center at x = +2, and $(y + 5)^2$, which is $(y - (-5))^2$, places it at y = −5. Read the center by asking what makes each squared term zero.

The right side is the radius squared. $x^2 + y^2 = 16$ has radius 4, not 16. Take the square root every time; a radius-16 circle would carry 256 on the right.

Equations often arrive expanded, and completing the square restores the standard form. The one rule: whatever you add inside, add to the other side too, including any coefficient multiplying the group.

Worked: $x^2 + y^2 - 6x + 4y - 3 = 0$. Group: $(x^2 - 6x) + (y^2 + 4y) = 3$. Complete each: add 9 and 4 to both sides: $(x - 3)^2 + (y + 2)^2 = 3 + 9 + 4 = 16$. A circle, center (3, −2), radius 4. Dropping the compensation leaves 3 on the right and a radius of √3: same algebra moves, wrong circle.

Coefficients raise the stakes: in $x^2 + 4(y^2 + 2y)$, completing inside the parentheses adds 1 inside but $4 \times 1 = 4$ to the other side, because the 4 multiplies everything in the group.

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the axes at $(\pm a, 0)$ and $(0, \pm b)$: the intercepts come from the square roots of the denominators. A hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ opens along the axis of its positive term, left-and-right here, up-and-down when the $y^2$ term is the positive one.

Circles, ellipses, and these hyperbolas all fail the vertical line test, and none of them care: they are implicit curves, valid exactly as in Topic 4.5. Of the family, only a parabola in the form $y = a(x-h)^2 + k$ has y as a function of x. Failing the test never demotes a conic to "not a real graph."