Section 1
Introduction to Conic Sections
Plot $x^2+y^2=1$ by solving for $y$ in terms of $x$. Enter the two solutions as $y_1$ and $y_2$. (One is a positive square root and the other is a negative square root.) Use a friendly window that includes the integers as grid points and has equal scales on the two axes. Based on the Pythagorean theorem, explain why the graph is the unit circle in Figure 12-1a.
Plot $4 x^2+9 y^2=36$ by first solving for $y$ in terms of $x$. Show that the result is the ellipse in Figure 12-1b.
The ellipse in Problem 2 is a dilation of the unit circle by a factor of 3 in the $x$-direction and by a factor of 2 in the $y$-direction. By making the right side equal 1 , transform the given equation to this equivalent form:$$\left(\frac{x}{3}\right)^2+\left(\frac{y}{2}\right)^2=1$$This form is sometimes called the standard form of the equation of the ellipse. Where do the two dilations show up in the transformed equation?
Plot $x^2-y^2=1$ by solving for $y$ in terms of $x$. Show that the result is the hyperbola in Figure 12-1c. Plot the two lines $y=x$ and $y=-x$. How are these lines related to the graph?
Plot the hyperbola $4 x^2-9 y^2=36$. Use a friendly window with an $x$-range of about $[-10,10]$, and use equal scales on the two axes. Show that the asymptotes now have slopes of $\pm \frac{2}{3}$ instead of \pm 1 .
Transform the equation in Problem 5 to make the right side equal 1, as in Problem 3. Show that the hyperbola in Problem 5 is a dilation of the hyperbola in Figure 12-1c with an $x$-dilation of 3 and a $y$-dilation of 2 . Tell where these dilation factors appear in the transformed equation.
The equation $x+y^2=1$ has only one squared term. Solve the equation for $y$ in terms of $x$ and plot the two solutions as $y_1$ and $y_2$. Show that the graph is the parabola in Figure 12-1d.
How could you tell from the equation before it is transformed whether its graph will be a circle, an ellipse, a hyperbola, or a parabola?