Ellipses and Hyperbolas
Definition of Ellipse
The locus of points P such that the sum of the distances from P to two fixed points is a constant is called an ellipse. Thus, let $ F_{1} $ and $ F_{2} $ be the two points (called foci, the plural of focus), then the defining relation for the ellipse is $ PF_{1} + PF_{2} = 2a $ . The line through the foci is called the focal axis of the ellipse; the point on the focal axis halfway between the foci is called the center; the points where the ellipse crosses the focal axis are called the vertices. The line segment joining the two vertices is called the major axis, and the line segment through the center, perpendicular to the major axis, with both endpoints on the ellipse, is called the minor axis. (See Fig. 38-1.)
An ellipse with focal axis parallel to one of the coordinate axes is said to be in standard orientation. If, in addition, the center of the ellipse is at the origin, the ellipse is said to be in one of two standard positions: with foci on the x-axis or with foci on the y-axis.
Graphs of Ellipses in Standard Position
Graphs of ellipses in standard position with their equations and characteristics are shown in the following table:
| FOCI ON x-AXIS | FOCI ON y-AXIS |
| Equation: $ + = 1 $ where $ b^{2} = a^{2} - c^{2} $ Note: $ a > b, a > c $ | Equation: $ + = 1 $ where $ b^{2} = a^{2} - c^{2} $ Note: $ a > b, a > c $ |
| Foci: $ F_{1}(-c, 0), F_{2}(c, 0) $ Vertices: $ (-a, 0), (a, 0) $ Center: $ (0, 0) $ | Foci: $ F_{1}(0, -c), F_{2}(0, c) $ Vertices: $ (0, -a), (0, a) $ Center: $ (0, 0) $ |
| FOCION x-AXIS | FOCION y-AXIS |
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| Figure 38-2 | Figure 38-3 |
Definition of Hyperbola
The locus of points P such that the absolute value of the difference of the distances from P to two fixed points is a constant is called a hyperbola. Thus, let $ F_{1} $ and $ F_{2} $ be the two points (foci), then the defining relation for the hyperbola is $ |PF_{1}-PF_{2}|=2a $ . The line through the foci is called the focal axis of the hyperbola; the point on the focal axis halfway between the foci is called the center; the points where the hyperbola crosses the focal axis are called the vertices. The line segment joining the two vertices is called the transverse axis. (See Fig. 38-4.)
A hyperbola with focal axis parallel to one of the coordinate axes is said to be in standard orientation. If, in addition, the center of the hyperbola is at the origin, the hyperbola is said to be in one of two standard positions: with foci on the x-axis or with foci on the y-axis.
Graphs of Hyperbolas in Standard Position
Graphs of hyperbolas in standard position with their equations and characteristics are shown in the following table:
| FOCION x-AXIS | FOCION y-AXIS |
| Foci: $ F_{1}(-c,0), F_{2}(c,0) $ Vertices: $ (-a,0), (a,0) $ Center: $ (0,0) $ | Foci: $ F_{1}(0,-c), F_{2}(0,c) $ Vertices: $ (0,-a), (0,a) $ Center: $ (0,0) $ |
| Equation: $ - = 1 $ where $ b^{2} = c^{2} - a^{2} $ Note: c > a, c > b | Equation: $ - = 1 $ where $ b^{2} = c^{2} - a^{2} $ Note: c > a, c > b |
| Asymptotes: $ y = x $ | Asymptotes: $ y = x $ |
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Definition of Eccentricity
A measure of the shape for an ellipse or hyperbola is the quantity $ e = $ , called the eccentricity. For an ellipse, 0 < e < 1, for a hyperbola e > 1.
SOLVED PROBLEMS
38.1. Derive the equation of an ellipse in standard position with foci on the x-axis.
Let \(P(x, y)\) be an arbitrary point on the ellipse. Given that the foci are \(F_{1}(-c,0)\) and \(F_{2}(c,0)\), then the definition of the ellipse \(PF_{1} + PF_{2} = 2a\) yields:
\[ \sqrt{(x+c)^{2}+(y-0)^{2}}+\sqrt{(x-c)^{2}+(y-0)^{2}}=2a \]
Subtracting one of the square roots from both sides, squaring, and simplifying yields:
\[ \sqrt{(x+c)^{2}+(y-0)^{2}}=2a-\sqrt{(x-c)^{2}+(y-0)^{2}} \]
\[ (x+c)^{2}+y^{2}=4a^{2}-4a\sqrt{(x-c)^{2}+(y-0)^{2}}+(x-c)^{2}+y^{2} \]
\[ \begin{aligned}x^{2}+2xc+c^{2}+y^{2}&=4a^{2}-4a\sqrt{(x-c)^{2}+(y-0)^{2}}+x^{2}-2xc+c^{2}+y^{2}\\4xc-4a^{2}&=-4a\sqrt{(x-c)^{2}+(y-0)^{2}}\\xc-a^{2}&=-a\sqrt{(x-c)^{2}+(y-0)^{2}}\end{aligned} \]
Now square both sides again and simplify:
\[ x^{2}c^{2}-2xca^{2}+a^{4}=a^{2}[(x-c)^{2}+y^{2}] \]
\[ x^{2}c^{2}-2xca^{2}+a^{4}=a^{2}x^{2}-2a^{2}xc+a^{2}c^{2}+a^{2}y^{2} \]
\[ x^{2}c^{2}-a^{2}x^{2}-a^{2}y^{2}=a^{2}c^{2}-a^{4} \]
\[ x^{2}(c^{2}-a^{2})-a^{2}y^{2}=a^{2}(c^{2}-a^{2}) \]
By the triangle inequality, the sum of two sides of a triangle is always greater than the third side. Hence (see Fig. 38-1)
\[ \begin{aligned}PF_{_{1}}+PF_{_{2}}&>F_{_{1}}F_{_{2}}\\2a&>2c\\a^{2}&>c^{2}\end{aligned} \]
Thus the quantity $ a{2}-c{2} $ must be positive. Set $ a{2}-c{2}=b^{2} $ . Then $ c{2}-a{2}=-b^{2} $ and the equation of the ellipse becomes:
\[ -b^{2}x^{2}-a^{2}y^{2}=-a^{2}b^{2} \]
\[ b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2} \]
This is generally written in standard form:
\[ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \]
Note that it follows from $ a{2}-c{2}=b^{2} $ that a>b.
38.2. Analyze the equation $ + = 1 $ of an ellipse in standard position, foci on the x-axis.
Set x = 0, then $ = 1 $ ; thus, $ y = b $ . Hence $ b $ are the y-intercepts.
Set \(y=0\), then \(\frac{x^{2}}{a^{2}}=1\); thus \(x=\pm a\). Hence \(\pm a\) are the \(x\)-intercepts.
Substitute -y for y: $ + = 1 $ ; $ + = 1 $ . Since the equation is unchanged, the graph has x-axis symmetry.
Substitute -x for \(x\): \(\frac{(-x)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\); \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\). Since the equation is unchanged, the graph has y-axis symmetry. It follows that the graph also has origin symmetry.
Note further that solving for y in terms of x yields $ y = $ ; hence $ -a x a $ for y to be real.
Similarly, \(-b \leq y \leq b\) for \(x\) to be real.
Summarizing, the graph is confined to the region between the intercepts $ a $ on the x-axis and $ b $ on the y-axis, and has all three symmetries. The graph of the ellipse is shown in Fig. 38-2.
38.3. Analyze the equation $ + = 1 $ of an ellipse in standard position, foci on the y-axis.
Set x = 0, then $ = 1 $ ; thus $ y = a $ . Hence $ a $ are the y-intercepts.
Set y = 0, then $ = 1 $ ; thus $ x = b $ . Hence $ b $ are the x-intercepts.
Substitute -y for y: $ + = 1 $ ; $ + = 1 $ . Since the equation is unchanged, the graph has x-axis symmetry.
Substitute -x for $ x:+=1;+=1 $ . Since the equation is unchanged, the graph has y-axis symmetry. It follows that the graph also has origin symmetry.
Note further that solving for y in terms of x yields $ y = $ ; hence $ -b x b $ for y to be real. Similarly $ -a y a $ for x to be real.
Summarizing, the graph is confined to the region between the intercepts $ b $ on the x-axis and $ a $ on the y-axis, and has all three symmetries. The graph of the ellipse is shown in Fig. 38-3.
38.4. Analyze and sketch graphs of the ellipses (a) $ 4x^{2} + 9y^{2} = 36 $ ; (b) $ 4x^{2} + y^{2} = 36 $ .
Written in standard form, the equation becomes
Written in standard form, the equation becomes becomes
\[ \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \]
\[ \frac{x^{2}}{9}+\frac{y^{2}}{36}=1 \]
Thus a = 3, b = 2.
Thus a = 6, b = 3.
Therefore $ c = = = . $
Therefore $ c = = = 3. $
Hence the ellipse is in standard position with foci at \((\pm\sqrt{5},0)\) on the \(x\)-axis, \(x\)-intercepts \((\pm3,0)\), and \(y\)-intercepts \((0,\pm2)\). The graph is shown in Fig. 38-7.
Hence the ellipse is in standard position with foci at \((0,\pm3\sqrt{3})\) on the \(y\)-axis, \(x\)-intercepts \((\pm3,0)\), and \(y\)-intercepts \((0,\pm6)\). The graph is shown in Fig. 38-8.
38.5. Derive the equation of a hyperbola in standard position with foci on the x-axis.
Let \(P(x,y)\) be an arbitrary point on the hyperbola. Given that the foci are \(F_{1}(-c,0)\) and \(F_{2}(c,0)\), then the definition of the hyperbola \(\left|PF_{1}-PF_{2}\right|=2a\); that is, \(PF_{1}-PF_{2}=\pm2a\) yields:
\[ \sqrt{(x+c)^{2}+(y-0)^{2}}-\sqrt{(x-c)^{2}+(y-0)^{2}}=\pm2a \]
Adding the second square root to both sides, squaring, and simplifying yields:
\[ \sqrt{(x+c)^{2}+(y-0)^{2}}=\pm2a+\sqrt{(x-c)^{2}+(y-0)^{2}} \]
\[ (x+c)^{2}+y^{2}=4a^{2}\pm4a\sqrt{(x-c)^{2}+(y-0)^{2}}+(x-c)^{2}+y^{2} \]
\[ x^{2}+2xc+c^{2}+y^{2}=4a^{2}\pm4a\sqrt{(x-c)^{2}+(y-0)^{2}}+x^{2}-2xc+c^{2}+y^{2} \]
\[ 4xc-4a^{2}=\pm4a\sqrt{(x-c)^{2}+(y-0)^{2}} \]
\[ xc-a^{2}=\pm a\sqrt{(x-c)^{2}+(y-0)^{2}} \]
Now square both sides again and simplify:
\[ x^{2}c^{2}-2xca^{2}+a^{4}=a^{2}[(x-c)^{2}+y^{2}] \]
\[ x^{2}c^{2}-2xca^{2}+a^{4}=a^{2}x^{2}-2a^{2}xc+a^{2}c^{2}+a^{2}y^{2} \]
\[ x^{2}c^{2}-a^{2}x^{2}-a^{2}y^{2}=a^{2}c^{2}-a^{4} \]
\[ x^{2}(c^{2}-a^{2})-a^{2}y^{2}=a^{2}(c^{2}-a^{2}) \]
By the triangle inequality, the sum of two sides of a triangle is always greater than the third side. Hence (see Fig. 38-4)
\[ \begin{aligned}PF_{2}+F_{1}F_{2}&>PF_{1}\\F_{1}F_{2}&>PF_{1}-PF_{2}\\2c&>2a\\c&>a\end{aligned} \]
Thus the quantity $ c{2}-a{2} $ must be positive. Set $ c{2}-a{2}=b^{2} $ . Then the equation of the hyperbola becomes:
\[ b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} \]
This is generally written in standard form:
\[ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \]
Note that it follows from $ c{2}-a{2}=b^{2} $ that c>b and c>a.
38.6. Analyze the equation $ - = 1 $ of a hyperbola in standard position, foci on the x-axis.
Set x = 0, then $ = -1 $ ; thus $ y^{2} = -b^{2} $ . Hence there can be no y-intercepts.
Set y = 0, then $ = 1 $ ; thus $ x = a $ . Hence $ a $ are the x-intercepts.
Substitute -y for y: $ - = 1 $ ; $ - = 1 $ . Since the equation is unchanged, the graph has x-axis symmetry.
Substitute -x for \(x:\frac{(-x)^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1;\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\). Since the equation is unchanged, the graph has y-axis symmetry. It follows that the graph also has origin symmetry.
Note further that solving for y in terms of x yields $ y = $ ; hence $ x a $ or $ x -a $ for y to be real.
Solving for x in terms of y yields $ x = $ ; hence y can take on any value.
It is left as an exercise to show that as x becomes arbitrarily large, the distance between the graphs of $ y = $ and the lines $ y = x $ becomes arbitrarily small, thus the lines are oblique asymptotes for the graph.
To draw the graph of the hyperbola, mark the intercepts $ a $ on the x-axis. Mark the points $ b $ on the y-axis. Draw vertical line segments through the points $ x = a $ and horizontal line segments through the points $ y = b $ to form the box shown in Fig. 38-5. Draw the diagonals of the box; these are the asymptotes of the hyperbola. Then sketch the hyperbola starting from the intercept x = a and approaching the asymptote y = bx/a. The remainder of the hyperbola follows from the symmetry with respect to axes and origin, as shown in Fig. 38-5.
38.7. Analyze the equation $ - = 1 $ of a hyperbola in standard position, foci on the y-axis.
Set \(x=0\), then \(\frac{y^{2}}{a^{2}}=1\); thus \(y=\pm a\). Hence \(\pm a\) are the \(y\)-intercepts.
Set y = 0, then $ - = 1 $ ; thus $ x^{2} = -b^{2} $ . Hence there can be no x-intercepts.
Substitute -y for y: $ - = 1 $ ; $ - = 1 $ . Since the equation is unchanged, the graph has x-axis symmetry.
Substitute -x for \(x\): \(\frac{y^{2}}{a^{2}} - \frac{(-x)^{2}}{b^{2}} = 1\); \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\). Since the equation is unchanged, the graph has y-axis symmetry. It follows that the graph also has origin symmetry.
Note further that solving for y in terms of x yields $ y = $ ; hence x can take on any value.
Solving for x in terms of y yields $ x = $ ; hence $ y a $ or $ y -a $ for x to be real.
It is left as an exercise to show that as x becomes arbitrarily large, the distance between the graphs of $ y = $ and the lines $ y = x $ becomes arbitrarily small, thus the lines are oblique asymptotes for the graph.
To draw the graph of the hyperbola, mark the intercepts $ a $ on the y-axis. Mark the points $ b $ on the x-axis. Draw vertical line segments through the points $ x = b $ and horizontal line segments through the points $ y = a $ to form the box shown in Fig. 38-6. Draw the diagonals of the box; these are the asymptotes of the hyperbola. Then sketch the hyperbola starting from the intercept y = a and approaching the asymptote y = ax/b. The remainder of the hyperbola follows from the symmetry with respect to axes and origin, as shown in Fig. 38-6.
38.8. Analyze and sketch graphs of the hyperbolas (a) $ 4x^{2} - 9y^{2} = 36 $ ; (b) $ y^{2} - 4x^{2} = 36 $ .
Written in standard form, the equation becomes
Written in standard form, the equation becomes
\[ \frac{x^{2}}{9}-\frac{y^{2}}{4}=1 \]
\[ \frac{y^{2}}{36}-\frac{x^{2}}{9}=1 \]
Thus a = 3, b = 2.
Therefore $ c = = = $ . Hence the hyperbola is in standard position with foci at $ (, 0) $ on the x-axis, x-intercepts $ (, 0) $ , and asymptotes $ y = x $ . The graph is shown in Fig. 38-9.
Thus a=6,b=3.
Therefore $ c = = = 3 $ . Hence the hyperbola is in standard position with foci at $ (0, ) $ on the y-axis, y-intercepts $ (0, ) $ , and asymptotes $ y = 2x $ . The graph is shown in Fig. 38-10.
38.9. Show in a table the characteristics of hyperbolas and ellipses in standard orientation, with center at the point $ (h,k) $ .
Shifting the center of the curves from the origin to the point $ (h, k) $ is reflected in the equations by replacing x with x - h and y with y - k, respectively. Hence the shifted curves can be described as follows:
| Ellipse; equation | Ellipse; equation | Hyperbola; equation | Hyperbola; equation |
| $ +=1 \(</td><td>\) +=1 \(</td><td>\) -=1 \(</td><td>\) -=1 $ | |||
| Foci: $ (hc,k) $ , Vertices: $ (ha,k) $ Endpoints of minor axis: $ (h,kb) $ | Foci: $ (h,kc) $ , Vertices: $ (h,ka) $ Endpoints of minor axis: $ (hb,k) $ | Foci: $ (hc,k) $ , Vertices: $ (ha,k) $ Asymptotes: $ (y-k)=(x-h) $ | Foci: $ (h,kc) $ , Vertices: $ (h,ka) $ Asymptotes: $ (y-k)=(x-h) $ |
38.10. Analyze and sketch the graph of $ 9x^{2} + 4y^{2} - 18x + 8y = 23 $ .
Complete the square on x and y.
\[ \begin{aligned}9(x^{2}-2x)+4(y^{2}+2y)&=23\\9(x^{2}-2x+1)+4(y^{2}+2y+1)&=23+9\cdot1+4\cdot1\\9(x-1)^{2}+4(y+1)^{2}&=36\\\frac{(x-1)^{2}}{4}+\frac{(y+1)^{2}}{9}&=1\end{aligned} \]
Comparing with the table in Problem 38.9, we find that this is the equation of an ellipse with center at $ (1,-1) $ . Since a > b, $ a^{2} = 9 $ and $ b^{2} = 4 $ , thus a = 3, b = 2, and $ c = = $ ; the focal axis is parallel to the y-axis. Foci: $ (h, k c) = (1, -1 ) $ . Vertices: $ (h, k a) = (1, -1 ) $ , thus $ (1, 2) $ and $ (1, -4) $ . Endpoints of minor axes: $ (h b, k) = (1 , -1) $ , thus $ (3, -1) $ and $ (-1, -1) $ . The graph is shown in Fig. 38-11.
38.11. Analyze and sketch the graph of $ 9x{2}-16y{2}-36x+32y=124 $ .
Complete the square on x and y.
\[ \begin{aligned}9(x^{2}-4x)-16(y^{2}-2y)&=124\\9(x^{2}-4x+4)-16(y^{2}-2y+1)&=124+9\cdot4-16\cdot1\\9(x-2)^{2}-16(y-1)^{2}&=144\\\frac{(x-2)^{2}}{16}-\frac{(y-1)^{2}}{9}&=1\end{aligned} \]
Comparing with the table in Problem 38.9, we find that this is the equation of a hyperbola with center at $ (2,1) $ . Since the coefficient of the square involving x is positive, the focal axis is parallel to the x-axis. (Note: For a hyperbola there is no restriction that a > b.) Hence $ a^{2} = 16 $ and $ b^{2} = 9 $ ; thus, a = 4, b = 3, and $ c = = = 5 $ . Foci: $ (h c, k) = (2 , 1) $ , thus, $ (7,1) $ and $ (-3,1) $ . Vertices: $ (h a, k) = (2 , 1) $ , thus, $ (6,1) $ and $ (-2,1) $ . Asymptotes: $ (y - k) = (x - h) $ , thus, $ (y - 1) = (x - 2) $ . Draw vertical lines through the vertices and horizontal lines through the points $ (h, k b) = (2, 1 ) $ , thus $ (2,4) $ and $ (2,-2) $ . These form the box. Sketch in the asymptotes and the diagonals of the box, then draw the hyperbola from the vertices out toward the asymptotes. The graph is shown in Fig. 38-12.
38.12. Analyze the quantity eccentricity e = c/a for an ellipse and a hyperbola.
For an ellipse, 0 < c < a, hence 0 < c/a = e < 1. The eccentricity measures the shape of the ellipse as follows:
If e is small, that is, close to 0, then c is small compared to a; hence, $ b = $ is close to a. Then the minor and major axes of the ellipse are roughly equal in size and the ellipse resembles a circle (the word eccentricity means departure from the center).
If e is large, that is, close to 1, then c is roughly equal to a; hence $ b = $ is close to 0. Then the major axis of the ellipse is substantially larger than the minor axis, and the ellipse has an elongated shape.
For a hyperbola, c > a; hence c/a = e > 1. The eccentricity measures the shape of the hyperbola by constraining the slope of the asymptotes, as follows:
If e is small, that is, close to 1, then c is roughly equal to a; hence $ b = $ is close to 0. Then the asymptotes, having slopes $ b/a $ or $ a/b $ , will seem close to the axes on which the vertices lie and the hyperbola will have a hairpin shape.
If e is large, then a is small compared to c; hence $ b = $ is close to c, and thus also large compared to a. Then the asymptotes will seem far from the axes on which the vertices lie and the hyperbola will seem wide.
38.13. Find the equation of an ellipse (a) in standard position with foci (±3,0) and y-intercepts (0,±2);
in standard orientation with foci $ (1,5) $ and $ (1,7) $ and eccentricity $ $
The ellipse is in standard position with foci on the x-axis. Hence it has an equation of the form $ + = 1 $ . From the position of the foci, c = 3; from the position of the y-intercepts, b = 2; hence $ a = = = $ . Thus the equation of the ellipse is $ + = 1 $ .
The center of the ellipse is midway between the foci, thus, at (1,6). Comparing with the table in Problem 38.9, the ellipse has an equation of the form $ + = 1 $ , with $ (h,k) = (1,6) $ . The distance between the foci = 2c = 2, thus c = 1. Since e = c/a = 1/2, it follows that a = 2 and $ b = = = $ . Thus the equation of the ellipse is $ + = 1 $ .
38.14. Find the equation of a hyperbola (a) in standard position with foci $ (,0) $ and x-intercepts $ (,0) $ ; (b) in standard orientation with foci $ (1,5) $ and $ (1,7) $ and eccentricity 2.
The hyperbola is in standard position with foci on the x-axis. Hence it has an equation of the form $ - = 1 $ . From the position of the foci, c = 3, from the position of the vertices, a = 2; hence $ b = = = $ . Thus the equation of the hyperbola is $ - = 1 $ .
The center of the hyperbola is midway between the foci, thus, at (1,6). Comparing with the table in Problem 38.9, we find that the hyperbola has an equation of the form $ - = 1 $ , with $ (h,k) = (1,6) $ . The distance between the foci = 2c = 2, thus c = 1. Since e = c/a = 2, it follows that a = 1/2 and $ b = = = ()/2 $ . Thus the equation of the hyperbola is $ - = 1 $ .
SUPPLEMENTARY PROBLEMS
38.15. Analyze and sketch graphs of the ellipses (a) $ + = 1 $ ; (b) $ 25x^{2} + 16y^{2} + 100x - 96y = 156 $ .
Ans. (a) Standard position, foci on x-axis at $ (,0) $ , vertices $ (,0) $ , endpoints of minor axis $ (0,) $ . See Fig. 38-13.
- Standard orientation, focal axis parallel to y-axis, center at $ (-2,3) $ , foci $ (-2,0) $ and $ (-2,6) $ , vertices $ (-2,-2) $ and $ (-2,8) $ , endpoints of minor axis $ (2,3) $ and $ (-6,3) $ . See Fig. 38-14.
38.16. Analyze and sketch graphs of the hyperbolas (a) $ -=1 $ ; (b) $ x{2}-y{2}+6x+34=0 $
Ans. (a) Standard position, foci on x-axis at $ (,0) $ , vertices $ (,0) $ , asymptotes $ y=x $ . See Fig. 38-15.
- Standard orientation, focal axis parallel to y-axis, foci at $ (-3, ) $ , vertices $ (-3, ) $ , asymptotes $ y = (x + 3) $ . See Fig. 38-16.
38.17. Show that as x becomes arbitrarily large, the distance between the graphs of $ y = $ and the lines $ y = x $ becomes arbitrarily small, thus the lines are oblique asymptotes for the graph.
38.18. Show that as x becomes arbitrarily large, the distance between the graphs of $ y = $ and the lines $ y = x $ becomes arbitrarily small, thus the lines are oblique asymptotes for the graph.
38.19. Find the eccentricity for each of the following:
\[ \begin{aligned}(a)\frac{x^{2}}{9}+\frac{y^{2}}{5}&=1;(b)25x^{2}+16y^{2}+100x-96y=156;(c)\frac{x^{2}}{9}-\frac{y^{2}}{5}=1;(d)x^{2}-y^{2}+6x+34=0.\end{aligned} \]
Ans. (a) 2/3; (b) 3/5; (c) $ /3 $ ; (d) $ $
38.20. Find the equation of an ellipse (a) with major vertices $ (,0) $ and eccentricity $ $ ; (b) with minor vertices $ (-3,4) $ and $ (1,4) $ and eccentricity $ $ .
Ans. (a) $ + = 1 $ ; (b) $ + = 1 $
38.21. Find the equation of a hyperbola (a) with vertices $ (0, ) $ and asymptotes $ y = 3x $ ;
- with foci (3, 6) and (11, 6) and eccentricity $ $
Ans. (a) $ -=1 $ ; (b) $ -=1 $
38.22. Use the definition of an ellipse $ PF_{1} + PF_{2} = 2a $ directly to find the equation of an ellipse with foci at $ (0,0) $ and $ (4,0) $ and major axis 2a = 6.
Ans. $ +=1 $