Determine what each equation represents: 1^° x^2-4y^2=0…

Determine what each equation represents: $1^\circ$ $x^2-4y^2=0$; $2^\circ$ $x^2+y^2+2x-4y+5=0$; $3^\circ$ $x^2+y^2-4x+6y+14=0$; $4^\circ$ $x^2+y^2-3x+3y-4=0$; $5^\circ$ $4x^2-9y^2-8x-36y-68=0$; $6^\circ$ $y^2-4y-4x+8=0$; $7^\circ$ $x^2-6x+9=0$.

Explanation of Method

Completing the square identifies seven curves, including degenerate and imaginary cases.

Step 1: $1^\circ$ Pair of lines

\[ (x-2y)(x+2y)=0\ \Longrightarrow\ y=\tfrac12x\ \text{ or }\ y=-\tfrac12x. \]

Step 2: $2^\circ$ One point

\[ (x+1)^2+(y-2)^2=0\ \Longrightarrow\ \text{only the point }(-1,2). \]

Step 3: $3^\circ$ Empty set

\[ (x-2)^2+(y+3)^2=-1\ \Longrightarrow\ \text{no real points (an “imaginary circle”).} \]

Step 4: $4^\circ$ Circle

\[ \left(x-\tfrac32\right)^2+\left(y+\tfrac32\right)^2=4+\tfrac94+\tfrac94=\tfrac{17}{2}. \]

\[\text{circle: centre }\left(\tfrac32,-\tfrac32\right),\quad r=\sqrt{\tfrac{17}{2}}=\frac{\sqrt{34}}{2}\approx2.92\]

Step 5: $5^\circ$ Hyperbola

\[ 4(x-1)^2-9(y+2)^2=68+4-36=36\ \Longrightarrow\ \frac{(x-1)^2}{9}-\frac{(y+2)^2}{4}=1. \]

Centre $(1,-2)$, $a=3$, $b=2$; asymptotes $y+2=\pm\tfrac23(x-1)$.

Step 6: $6^\circ$ Parabola

\[ (y-2)^2=4x-4=4(x-1). \]

Vertex $(1,2)$, $2p=4$, so $p=2$; focus $(2,2)$, directrix $x=0$.

Step 7: $7^\circ$ Double line

\[ (x-3)^2=0\ \Longrightarrow\ \text{the line }x=3\text{ counted twice.} \]

	Type	Key data
$1^\circ$	two intersecting lines	$y=\pm x/2$
$2^\circ$	point	$(-1,2)$
$3^\circ$	empty set	$r^2=-1$
$4^\circ$	circle	$\left(\tfrac32,-\tfrac32\right),\ r=\tfrac{\sqrt{34}}2$
$5^\circ$	hyperbola	$(1,-2),\ a=3,\ b=2$
$6^\circ$	parabola	vertex $(1,2),\ p=2$
$7^\circ$	double line	$x=3$

$1^\circ$ $7^\circ$ $2^\circ\,(-1,2)$

Degenerate cases: $1^\circ$ the pair of lines $y=\pm\tfrac x2$; $2^\circ$ the single point $(-1,2)$; $7^\circ$ the double line $x=3$.

$r=\tfrac{\sqrt{34}}2$ $(\tfrac32,-\tfrac32)$

$4^\circ$ Circle with centre $\left(\tfrac32,-\tfrac32\right)$ and $r=\tfrac{\sqrt{34}}2\approx2.92$ (not $\tfrac{5}{\sqrt2}$).

$(1,-2)$

$5^\circ$ Hyperbola $\dfrac{(x-1)^2}{9}-\dfrac{(y+2)^2}{4}=1$, asymptotes dashed.

$(1,2)$ $F$

$6^\circ$ Parabola $(y-2)^2=4(x-1)$: vertex $(1,2)$, focus $F(2,2)$, directrix $x=0$ (dashed).

Result

\[\boxed{\,\text{circle: centre }\left(\tfrac32,-\tfrac32\right),\quad r=\sqrt{\tfrac{17}{2}}=\frac{\sqrt{34}}{2}\approx2.92\,}\]

Remark

Correction: $4+\tfrac94+\tfrac94=\tfrac{16+9+9}{4}=\tfrac{34}{4}=\tfrac{17}{2}$, not $\tfrac{25}{2}$. So $r=\tfrac{\sqrt{34}}2$, not $\tfrac{5}{\sqrt2}$.