Ellipse and circle: three ways to obtain faithful sketches

Rewrite the denominators as $2^2$ and $3^2$: $\dfrac{(x-1)^2}{2^2}+\dfrac{[y-(-2)]^2}{3^2}=1$. The centre is $(1,-2)$, horizontal semi-axis $2$, vertical semi-axis $3$. The endpoints are $(-1,-2),(3,-2),(1,1),(1,-5)$. Draw a smooth closed ellipse, with symmetry lines $x=1$ and $y=-2$. The Teacher copy fills the centre, sketch and symmetry blanks in the Student copy.

The source’s margin guidance says to convert the equation to standard form and complete the square for the $x$ and $y$ terms separately. Thus $x^2-5x+y^2+y-\tfrac52=0$ becomes $(x-\tfrac52)^2-\tfrac{25}{4}+(y+\tfrac12)^2-\tfrac14-\tfrac52=0,$ so $\begin{aligned}&(x-\tfrac52)^2+(y+\tfrac12)^2\\ & =9\\ & =3^2.\end{aligned}$ It is a circle centred at $(\tfrac52,-\tfrac12)$ with radius $3$. Its extreme coordinates are $x=-\tfrac12,\tfrac{11}{2}$ and $y=-\tfrac72,\tfrac52$. Every straight line passing through its centre is a line of symmetry; the two coordinate-parallel lines are only two examples.

Follow the source’s calculator table: apps → [2:Conics] → [2:ELLIPSE] → [2:]. The selected form is $\dfrac{(X-H)^2}{B^2}+\dfrac{(Y-K)^2}{A^2}=1$, since the application assumes $A>B>0$. Enter A=3, B=2, H=1, K=−2, then graph. The displayed graph is the vertical ellipse from the algebraic method. Use trace to view coordinates, retaining the source warning that traced values may not be accurate. The labels A/B in this selected app form must not be swapped with the horizontal/vertical denominators.

Follow apps → [2:Conics] → [1:CIRCLE] → [2:]. This form is $AX^2+AY^2+BX+CY+D=0$. Enter A=1, B=−5, C=1, D=−5/2, then graph and trace if needed. The source also offers [1:], the centre-radius form $(X-H)^2+(Y-K)^2=R^2$, corresponding to $(x-\tfrac52)^2+(y+\tfrac12)^2=9$. Its remark recommends using the given equation directly as a check against algebraic manipulation errors. Traced coordinates may not be accurate; the exact centre and radius come from the algebra.

Press y= and enter $Y_1=\sqrt{9-\tfrac94(x-1)^2}-2,\qquad Y_2=-\sqrt{9-\tfrac94(x-1)^2}-2.$ The real domain is $-1\le x\le3$. Press graph. The source’s first screen has unequal axis scales, so the displayed shape can be misleading. Press zoom → [5:ZSquare] to use equal axis units. The source’s final screen has small rendering gaps, but the mathematical ellipse and the answer drawn must be continuous and smooth: the two branches meet at $(-1,-2)$ and $(3,-2)$.

Press y= and enter $Y_1=\sqrt{9-(x-\tfrac52)^2}-\tfrac12,\qquad Y_2=-\sqrt{9-(x-\tfrac52)^2}-\tfrac12.$ The real domain is $-\tfrac12\le x\le\tfrac{11}{2}$. Press graph, then zoom → [5:ZSquare] to correct the unequal-axis distortion shown in the source. The two branches meet at $(-\tfrac12,-\tfrac12)$ and $(\tfrac{11}{2},-\tfrac12)$. The screen’s pixel gaps are not discontinuities of the circle: draw a smooth, continuous closed curve.