Practice 1 Solutions

Complete the sentence with always, sometimes, or never.

A hyperbola will sometimes intersect the larger (longer) axis.

Note

The hyperbola intersects the transverse axis.

Conic sections are sometimes functions.

Note

A parabola can be a function when it opens up or down.

A circle is always a special form of an ellipse.

Note

When the values in the denominator are equal $(a = b)$ , the ellipse is also a circle.

Mark the center, vertices and co-vertices, and asymptotes on the graph of $\frac{{(x + 4)}^{2}}{16} - \frac{{(y - 2)}^{2}}{25} = 1$ . Then write the equation of the asymptotes.

$Center : (- 4, 2) a = 4 b = 5 S lope of asymptotes : \pm \frac{5}{4}$

$y - 2 = \pm \frac{5}{4} (x + 4)$

Mark the center, vertices and co-vertices, and asymptotes on the graph of $\frac{y^{2}}{4} - \frac{{(x - 1)}^{2}}{9} = 1 .$ Then write the equation of the asymptotes.

$C enter : (1, 0) a = 3 b = 2 S lope of asymptotes : \pm \frac{2}{3}$

$y = \pm \frac{2}{3} (x - 1)$

Graph the hyperbola: $\frac{x^{2}}{9} - \frac{{(y - 3)}^{2}}{25} = 1$

$C enter : (0, 3) a = 3 b = 5 S lope of asymptote : \pm \frac{5}{3}$

Graph the hyperbola: $16 x^{2} - 64 x - 36 y^{2} + 72 y = 116$

$16 (x^{2} - 4 x) - 36 (y^{2} - 2 y) = 116 16 (x^{2} - 4 x + {(- \frac{4}{2})}^{2}) - 36 (y^{2} - 2 y + {(- \frac{2}{2})}^{2}) = 116 + 16 {(- \frac{4}{2})}^{2} - 36 {(- \frac{2}{2})}^{2} 16 {(x - 2)}^{2} - 36 {(y - 1)}^{2} = 116 + 64 - 36 16 {(x - 2)}^{2} - 36 {(y - 1)}^{2} = 144 \frac{{(x - 2)}^{2}}{9} - \frac{{(y - 1)}^{2}}{4} = 1 C enter : (2, 1) a = 3 b = 2 S lope of asymptotes : \pm \frac{2}{3}$

Graph: $\frac{y^{2}}{49} - \frac{x^{2}}{81} = 1$

$C enter : (0, 0) a = 7 b = 9 S lope of asymptotes : \pm \frac{7}{9}$

Graph the horizontal hyperbola with asymptotes: $y - 6 = \pm \frac{6}{7} (x + 1)$

$C enter : (- 1, 6) a = 7 b = 6 T ransverse horizontal axis \frac{{(x + 1)}^{2}}{49} - \frac{{(y - 6)}^{2}}{36} = 1$

Graph the hyperbola with vertices at $(- 8, 0)$ and $(8, 0)$ and co-vertices at $(0, 3)$ and $(0, - 3)$ .

$C enter : (0, 0) a = 8 b = 3 Horizontal \frac{x^{2}}{64} - \frac{y^{2}}{9} = 1$

Mike was asked to write the hyperbola $9 x^{2} - 36 x - 16 y^{2} - 96 y = 252$ in standard form. Verify or find the error in his solution. If there is an error, determine the correct equation. 

$9 x^{2} - 36 x - 16 y^{2} - 96 y = 252 9 (x^{2} - 4 x) - 16 (y^{2} - 6 y) = 252 9 (x^{2} - 4 x + {(- \frac{4}{2})}^{2}) - 16 (y^{2} + 6 y + {(\frac{6}{2})}^{2}) = 252 + 9 {(- \frac{4}{2})}^{2} + 16 {(\frac{6}{2})}^{2} 9 {(x - 2)}^{2} - 16 {(y - 3)}^{2} = 432 \frac{{(x - 2)}^{2}}{48} - \frac{{(y - 3)}^{2}}{27} = 1$

$9 (x^{2} - 4 x + {(- \frac{4}{2})}^{2}) - 16 (y^{2} + 6 y + {(\frac{6}{2})}^{2}) = 252 + 9 {(- \frac{4}{2})}^{2} - 16 {(\frac{6}{2})}^{2} Mike should have subtracted 16 (9) . 9 {(x - 2)}^{2} - 16 {(y - 3)}^{2} = 144 \frac{{(x - 2)}^{2}}{16} - \frac{{(y - 3)}^{2}}{9} = 1$

Match the graph to the equation.

$x^{2} - 12 x + y^{2} - 2 y = - 1$

$A = C$

$x^{2} - y^{2} + 4 y = 13$

A and C have different signs.

$6 x^{2} + y^{2} + 4 y = 9$

$A \neq C$ and have the same sign.

Name the conic section. Explain.

$8 y + 2 - x^{2} = x - 1$

$A = - 1, C = 0$

Parabola

$3 y^{2} - 6 y = 3 x^{2} + 15$

$A = - 3, C = 3$
A and C have different signs.

Hyperbola

$4 x^{2} = - 9 y^{2} + 36 y$

$4 x^{2} + 9 y^{2} - 36 y = 0 A = 9, C = 4$

Ellipse

$x = 2 y^{2} + 4$

$A = 0, C = 2$

Parabola