Ellipses, Hyperbolas and Mixed Practice, Worked Solutions

Ques­tion 1: Read­ing off the parts of two ellipses#

The prob­lem#

Find the foci, ver­tices, eccen­tric­ity and latus rec­tum of $\displaystyle \frac{x^2}{49} + \frac{y^2}{36} = 1$ and of $16x^2 + y^2 = 16$.

Under­stand­ing the prob­lem#

You are given two equa­tions of ellipses. For each, you must find four things: the two foci, the two ver­tices (ends of the major axis), the eccen­tric­ity $e$, and the length of the latus rec­tum. The sec­ond equa­tion is not yet in stan­dard form, so you must first rewrite it with $1$ on the right.

The idea#

Put each equa­tion in the form $\displaystyle \frac{x^2}{\square} + \frac{y^2}{\square} = 1$. The larger denom­i­na­tor is $a^2$ and shows which axis is the major axis. Then use $c^2 = a^2 - b^2$, $\displaystyle e = \frac{c}{a}$ and latus rec­tum $\displaystyle \frac{2b^2}{a}$.

Step-by-step solu­tion#

Part (a): $\displaystyle \frac{x^2}{49} + \frac{y^2}{36} = 1$

Step 1. Iden­tify $a^2$ and $b^2$. The larger denom­i­na­tor, $49$, is under $x^2$, so the major axis lies along the $x$-axis.

$$a^2 = 49,\ a = 7, \qquad b^2 = 36,\ b = 6.$$

Step 2. Find $c$ from $c^2 = a^2 - b^2$.

$$c^2 = 49 - 36 = 13, \qquad c = \sqrt{13}.$$

Step 3. Write the foci and ver­tices. With the major axis on the $x$-axis, both lie on the $x$-axis.

$$\text{foci } (\pm\sqrt{13},\ 0), \qquad \text{vertices } (\pm 7,\ 0).$$

Step 4. Eccen­tric­ity and latus rec­tum.

$$\displaystyle e = \frac{c}{a} = \frac{\sqrt{13}}{7}, \qquad \text{latus rectum} = \frac{2b^2}{a} = \frac{2 \times 36}{7} = \frac{72}{7}.$$

Part (b): $16x^2 + y^2 = 16$

Step 1. Divide both sides by $16$ to get $1$ on the right.

$$\displaystyle \frac{16x^2}{16} + \frac{y^2}{16} = 1 \quad\Longrightarrow\quad \frac{x^2}{1} + \frac{y^2}{16} = 1.$$

Step 2. Now the larger denom­i­na­tor, $16$, is under $y^2$, so the major axis is ver­ti­cal (along the $y$-axis).

$$a^2 = 16,\ a = 4, \qquad b^2 = 1,\ b = 1.$$

Step 3. Find $c$.

$$c^2 = a^2 - b^2 = 16 - 1 = 15, \qquad c = \sqrt{15}.$$

Step 4. Foci and ver­tices lie on the $y$-axis.

$$\text{foci } (0,\ \pm\sqrt{15}), \qquad \text{vertices } (0,\ \pm 4).$$

Step 5. Eccen­tric­ity and latus rec­tum.

$$\displaystyle e = \frac{\sqrt{15}}{4}, \qquad \text{latus rectum} = \frac{2b^2}{a} = \frac{2 \times 1}{4} = \frac{1}{2}.$$

Check­ing the answer#

In both parts $e < 1$ (since $\sqrt{13} \approx 3.6 < 7$ and $\sqrt{15} \approx 3.9 < 4$), as it must be for an ellipse. The foci lie inside the ellipse, between the cen­tre and the ver­tices: $\sqrt{13} < 7$ and $\sqrt{15} < 4$.

Answer#

(a) Foci $(\pm\sqrt{13}, 0)$, ver­tices $(\pm 7, 0)$, $\displaystyle e = \tfrac{\sqrt{13}}{7}$, latus rec­tum $\displaystyle \tfrac{72}{7}$. (b) Foci $(0, \pm\sqrt{15})$, ver­tices $(0, \pm 4)$, $\displaystyle e = \tfrac{\sqrt{15}}{4}$, latus rec­tum $\displaystyle \tfrac{1}{2}$.

Com­mon mis­take to avoid#

In part (b), do not take $a^2 = 1$ just because $x^2$ comes first. Always com­pare the denom­i­na­tors: the larger one is $a^2$.

Ques­tion 2: Ellipse from ver­tices and eccen­tric­ity#

The prob­lem#

Find the ellipse with ver­tices $(0, \pm 10)$ and $\displaystyle e = \frac{3}{5}$.

Under­stand­ing the prob­lem#

The ver­tices are the ends of the major axis. Here they are on the $y$-axis, so the major axis is ver­ti­cal and the cen­tre is at the ori­gin. You also know the eccen­tric­ity. You must find the equa­tion.

The idea#

For a ver­ti­cal major axis the equa­tion is $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$. The ver­tices give $a$; the eccen­tric­ity gives $c = ae$; then $b^2 = a^2 - c^2$.

Step-by-step solu­tion#

Step 1. Ver­tices $(0, \pm 10)$ give the semi-major axis.

$$a = 10, \qquad a^2 = 100.$$

Step 2. Use $\displaystyle e = \frac{c}{a}$ to find $c$.

$$\displaystyle c = ae = 10 \times \frac{3}{5} = 6.$$

Step 3. Find $b^2$.

$$b^2 = a^2 - c^2 = 100 - 36 = 64.$$

Step 4. Put $a^2$ under $y^2$ (ver­ti­cal major axis) and $b^2$ under $x^2$.

$$\displaystyle \frac{x^2}{64} + \frac{y^2}{100} = 1.$$

Check­ing the answer#

Put $x = 0$: $y^2 = 100$, $y = \pm 10$, the given ver­tices. And $c = \sqrt{100 - 64} = 6$, so $\displaystyle e = \tfrac{6}{10} = \tfrac{3}{5}$. ✓

Answer#

$\displaystyle \frac{x^2}{64} + \frac{y^2}{100} = 1$

Ques­tion 3: Ellipse through two given points#

The prob­lem#

Find the ellipse with cen­tre at the ori­gin, major axis on the $x$-axis, through $(4, 3)$ and $(-1, 4)$.

Under­stand­ing the prob­lem#

The ellipse has the form $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, but you do not know $a^2$ or $b^2$. Each point on the ellipse must sat­isfy the equa­tion, which gives you two equa­tions in two unknowns.

The idea#

Sub­sti­tute each point into the stan­dard equa­tion. To make the alge­bra easy, treat $\displaystyle \frac{1}{a^2}$ and $\displaystyle \frac{1}{b^2}$ as the unknowns, the equa­tions are then lin­ear.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $(4, 3)$.

$$\displaystyle \frac{16}{a^2} + \frac{9}{b^2} = 1. \qquad (1)$$

Step 2. Sub­sti­tute $(-1, 4)$. Note $(-1)^2 = 1$.

$$\displaystyle \frac{1}{a^2} + \frac{16}{b^2} = 1. \qquad (2)$$

Step 3. Let $\displaystyle p = \frac{1}{a^2}$ and $\displaystyle q = \frac{1}{b^2}$. The equa­tions become

$$16p + 9q = 1, \qquad p + 16q = 1.$$

Step 4. From the sec­ond equa­tion, $p = 1 - 16q$. Sub­sti­tute into the first.

$$\displaystyle \begin{aligned} 16(1 - 16q) + 9q &= 1\\ 16 - 256q + 9q &= 1\\ -247q &= -15\\ q &= \frac{15}{247}. \end{aligned}$$

Step 5. Find $p$.

$$\displaystyle p = 1 - 16 \cdot \frac{15}{247} = \frac{247 - 240}{247} = \frac{7}{247}.$$

Step 6. Turn back into $a^2$ and $b^2$.

$$\displaystyle a^2 = \frac{1}{p} = \frac{247}{7}, \qquad b^2 = \frac{1}{q} = \frac{247}{15}.$$

So the ellipse is

$$\displaystyle \frac{7x^2}{247} + \frac{15y^2}{247} = 1, \quad\text{that is,}\quad 7x^2 + 15y^2 = 247.$$

Check­ing the answer#

At $(4, 3)$: $7(16) + 15(9) = 112 + 135 = 247$. ✓ At $(-1, 4)$: $7(1) + 15(16) = 7 + 240 = 247$. ✓ Also $\displaystyle a^2 = \tfrac{247}{7} \approx 35.3 > b^2 = \tfrac{247}{15} \approx 16.5$, so the major axis is indeed on the $x$-axis.

Answer#

$\displaystyle a^2 = \tfrac{247}{7}$, $\displaystyle b^2 = \tfrac{247}{15}$: the ellipse is $7x^2 + 15y^2 = 247$.

Ques­tion 4: Read­ing off the parts of two hyper­bo­las#

The prob­lem#

Find the foci, ver­tices, eccen­tric­ity and latus rec­tum of $\displaystyle \frac{y^2}{16} - \frac{x^2}{9} = 1$ and of $3x^2 - y^2 = 3$.

Under­stand­ing the prob­lem#

Both are hyper­bo­las (a minus sign between the squared terms). For each you need the foci, ver­tices, $e$ and latus rec­tum. The sec­ond must first be put in stan­dard form.

The idea#

For a hyper­bola, $a^2$ is the denom­i­na­tor of the pos­i­tive term, and that ter­m's vari­able gives the direc­tion of the trans­verse axis. Then $c^2 = a^2 + b^2$.

Step-by-step solu­tion#

Part (a): $\displaystyle \frac{y^2}{16} - \frac{x^2}{9} = 1$

Step 1. The $y^2$ term is pos­i­tive, so the trans­verse axis is ver­ti­cal.

$$a^2 = 16,\ a = 4, \qquad b^2 = 9,\ b = 3.$$

Step 2. For a hyper­bola, $c^2 = a^2 + b^2$.

$$c^2 = 16 + 9 = 25, \qquad c = 5.$$

Step 3. Foci and ver­tices on the $y$-axis.

$$\text{foci } (0,\ \pm 5), \qquad \text{vertices } (0,\ \pm 4).$$

Step 4. Eccen­tric­ity and latus rec­tum.

$$\displaystyle e = \frac{5}{4}, \qquad \text{latus rectum} = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}.$$

Part (b): $3x^2 - y^2 = 3$

Step 1. Divide by $3$.

$$\displaystyle \frac{x^2}{1} - \frac{y^2}{3} = 1.$$

Step 2. The $x^2$ term is pos­i­tive, so the trans­verse axis is hor­i­zon­tal.

$$a^2 = 1,\ a = 1, \qquad b^2 = 3,\ b = \sqrt{3}.$$

Here $b > a$, that is allowed for a hyper­bola.

Step 3. Find $c$.

$$c^2 = 1 + 3 = 4, \qquad c = 2.$$

Step 4. Foci, ver­tices, eccen­tric­ity, latus rec­tum.

$$\displaystyle \text{foci } (\pm 2,\ 0), \quad \text{vertices } (\pm 1,\ 0), \quad e = \frac{2}{1} = 2, \quad \text{latus rectum} = \frac{2 \times 3}{1} = 6.$$

Check­ing the answer#

Both eccen­tric­i­ties are greater than $1$, as they must be for hyper­bo­las. The foci lie far­ther from the cen­tre than the ver­tices ($5 > 4$ and $2 > 1$).

Answer#

(a) Foci $(0, \pm 5)$, ver­tices $(0, \pm 4)$, $\displaystyle e = \tfrac{5}{4}$, latus rec­tum $\displaystyle \tfrac{9}{2}$. (b) Foci $(\pm 2, 0)$, ver­tices $(\pm 1, 0)$, $e = 2$, latus rec­tum $6$.

Com­mon mis­take to avoid#

Do not use $c^2 = a^2 - b^2$ for a hyper­bola. For a hyper­bola it is $c^2 = a^2 + b^2$.

Ques­tion 5: Hyper­bola from ver­tices and eccen­tric­ity#

The prob­lem#

Find the hyper­bola with ver­tices $(\pm 7, 0)$ and $\displaystyle e = \frac{4}{3}$.

Under­stand­ing the prob­lem#

The ver­tices are on the $x$-axis, so the trans­verse axis is hor­i­zon­tal and the equa­tion has the form $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. You must find $a^2$ and $b^2$.

The idea#

Ver­tices give $a$. Then $c = ae$ and $b^2 = c^2 - a^2$.

Step-by-step solu­tion#

Step 1. From the ver­tices,

$$a = 7, \qquad a^2 = 49.$$

Step 2. Find $c$.

$$\displaystyle c = ae = 7 \times \frac{4}{3} = \frac{28}{3}.$$

Step 3. Find $b^2$. Write $49$ as $\displaystyle \tfrac{441}{9}$ to sub­tract.

$$\displaystyle b^2 = c^2 - a^2 = \frac{784}{9} - \frac{441}{9} = \frac{343}{9}.$$

Step 4. Write the equa­tion. Divid­ing by $\displaystyle \tfrac{343}{9}$ is the same as mul­ti­ply­ing by $\displaystyle \tfrac{9}{343}$.

$$\displaystyle \frac{x^2}{49} - \frac{y^2}{343/9} = 1 \quad\Longrightarrow\quad \frac{x^2}{49} - \frac{9y^2}{343} = 1.$$

Check­ing the answer#

$\displaystyle c^2 = a^2 + b^2 = \tfrac{441 + 343}{9} = \tfrac{784}{9}$, so $\displaystyle c = \tfrac{28}{3}$ and $\displaystyle e = \tfrac{28/3}{7} = \tfrac{4}{3}$. ✓

Answer#

$\displaystyle \frac{x^2}{49} - \frac{9y^2}{343} = 1$

Ques­tion 6: Hyper­bola from foci and latus rec­tum#

The prob­lem#

Find the hyper­bola with foci $(\pm 4, 0)$ and latus rec­tum $12$.

Under­stand­ing the prob­lem#

Foci on the $x$-axis mean the form $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $c = 4$. The latus rec­tum length gives a sec­ond con­di­tion. Two con­di­tions, two unknowns.

The idea#

Use $\displaystyle \frac{2b^2}{a} = 12$ together with $a^2 + b^2 = c^2 = 16$. Elim­i­nate $b^2$ to get a qua­dratic in $a$.

Step-by-step solu­tion#

Step 1. From the latus rec­tum,

$$\displaystyle \frac{2b^2}{a} = 12 \quad\Longrightarrow\quad b^2 = 6a.$$

Step 2. From the foci, $c = 4$, so

$$a^2 + b^2 = 16.$$

Step 3. Sub­sti­tute $b^2 = 6a$.

$$a^2 + 6a - 16 = 0.$$

Step 4. Fac­torise: we need two num­bers with prod­uct $-16$ and sum $6$, namely $8$ and $-2$.

$$(a + 8)(a - 2) = 0 \quad\Longrightarrow\quad a = 2 \ \text{ or } \ a = -8.$$

A length can­not be neg­a­tive, so $a = 2$.

Step 5. Find $b^2$ and write the equa­tion.

$$\displaystyle a^2 = 4, \qquad b^2 = 6a = 12, \qquad \frac{x^2}{4} - \frac{y^2}{12} = 1.$$

Check­ing the answer#

$a^2 + b^2 = 4 + 12 = 16 = c^2$ ✓, and latus rec­tum $\displaystyle \tfrac{2 \times 12}{2} = 12$ ✓.

Answer#

$\displaystyle \frac{x^2}{4} - \frac{y^2}{12} = 1$

Ques­tion 7: Iden­ti­fy­ing con­ics#

The prob­lem#

Iden­tify the conic: $x^2 + y^2 - 4x = 0$; $4x^2 + 9y^2 = 36$; $x^2 - 4y^2 = 4$; $y^2 = -6x$.

Under­stand­ing the prob­lem#

For each equa­tion, decide whether it is a cir­cle, an ellipse, a hyper­bola or a parabola.

The idea#

Look at the squared terms:

• both $x^2$ and $y^2$, equal coef­fi­cients, same sign → cir­cle;
• both, same sign, dif­fer­ent coef­fi­cients → ellipse;
• both, oppo­site signs → hyper­bola;
• only one vari­able squared → parabola.

Step-by-step solu­tion#

Step 1. $x^2 + y^2 - 4x = 0$. Both squares have coef­fi­cient $1$. Com­plet­ing the square con­firms it:

$$(x - 2)^2 + y^2 = 4,$$

a cir­cle with cen­tre $(2, 0)$, radius $2$.

Step 2. $4x^2 + 9y^2 = 36$. Same sign, dif­fer­ent coef­fi­cients. Divid­ing by $36$:

$$\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1,$$

an ellipse.

Step 3. $x^2 - 4y^2 = 4$. Oppo­site signs. Divid­ing by $4$:

$$\displaystyle \frac{x^2}{4} - \frac{y^2}{1} = 1,$$

a hyper­bola.

Step 4. $y^2 = -6x$. Only $y$ is squared, so it is a parabola (open­ing to the left, since $4a = -6$ is neg­a­tive).

Check­ing the answer#

Each equa­tion was rewrit­ten into a stan­dard form from the chap­ter, which con­firms the type.

Answer#

Cir­cle; ellipse; hyper­bola; parabola.

Ques­tion 8: Locus of a point on a slid­ing rod#

The prob­lem#

A rod of length $15$ cm slides with its ends on the axes. Find the locus of the point $P$ on the rod $5$ cm from the end on the $x$-axis.

Under­stand­ing the prob­lem#

One end $A$ of the rod stays on the $x$-axis and the other end $B$ on the $y$-axis, while the rod moves. The point $P$ is fixed on the rod, $5$ cm from $A$ (so $10$ cm from $B$). As the rod slides, $P$ traces a curve, its locus. You must find its equa­tion.

The idea#

Name the ends $A(\alpha, 0)$ and $B(0, \beta)$. The rod's length gives one equa­tion in $\alpha$ and $\beta$. Express $P$'s coor­di­nates in terms of $\alpha$ and $\beta$ using the sec­tion for­mula, then elim­i­nate $\alpha$ and $\beta$.

Step-by-step solu­tion#

Step 1. Length $AB = 15$ gives, by the dis­tance for­mula,

$$\alpha^2 + \beta^2 = 15^2 = 225. \qquad (1)$$

Step 2. $P$ divides $AB$ with $AP : PB = 5 : 10 = 1 : 2$. By the sec­tion for­mula,

$$\displaystyle P = \left(\frac{2 \cdot \alpha + 1 \cdot 0}{3},\ \frac{2 \cdot 0 + 1 \cdot \beta}{3}\right) = \left(\frac{2\alpha}{3},\ \frac{\beta}{3}\right).$$

Step 3. Let $P = (x, y)$. Then

$$\displaystyle x = \frac{2\alpha}{3} \Rightarrow \alpha = \frac{3x}{2}, \qquad y = \frac{\beta}{3} \Rightarrow \beta = 3y.$$

Step 4. Sub­sti­tute into $(1)$.

$$\displaystyle \begin{aligned} \left(\frac{3x}{2}\right)^2 + (3y)^2 &= 225\\ \frac{9x^2}{4} + 9y^2 &= 225\\ \frac{x^2}{4} + y^2 &= 25 \qquad (\text{divide by } 9)\\ \frac{x^2}{100} + \frac{y^2}{25} &= 1 \qquad (\text{divide by } 25). \end{aligned}$$

Check­ing the answer#

When the rod lies flat on the $x$-axis ($\alpha = 15$, $\beta = 0$), $P = (10, 0)$, and $\displaystyle \tfrac{100}{100} + 0 = 1$ ✓. When the rod stands on the $y$-axis ($\alpha = 0$, $\beta = 15$), $P = (0, 5)$, and $\displaystyle 0 + \tfrac{25}{25} = 1$ ✓.

Answer#

The locus is the ellipse $\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$.

Com­mon mis­take to avoid#

Take the ratio in the right order: $P$ is $5$ cm from $A$, so it is closer to $A$, and its $x$-coor­di­nate is $\displaystyle \tfrac{2}{3}$ of $\alpha$, not $\displaystyle \tfrac{1}{3}$.

Ques­tion 9: Height of an ellip­ti­cal arch#

The prob­lem#

An arch is a semi-ellipse $10$ m wide and $3$ m high at the cen­tre. Find its height $2$ m from one end.

Under­stand­ing the prob­lem#

Pic­ture the arch stand­ing on the ground. Its base is $10$ m wide and its high­est point is $3$ m above the mid­dle of the base. You need the height of the arch at a point on the ground $2$ m in from one end.

The idea#

Put the ori­gin at the mid­dle of the base with the $x$-axis along the ground. Then the arch is the upper half of an ellipse with $a = 5$ (half the width) and $b = 3$ (the height). A point $2$ m from an end is $5 - 2 = 3$ m from the cen­tre.

Step-by-step solu­tion#

Step 1. Write the ellipse.

$$\displaystyle a = \frac{10}{2} = 5, \quad b = 3, \qquad \frac{x^2}{25} + \frac{y^2}{9} = 1.$$

Step 2. The point $2$ m from the end at $x = 5$ has $x = 5 - 2 = 3$.

Step 3. Solve for $y$ (take the pos­i­tive root, since it is a height).

$$\displaystyle \begin{aligned} \frac{y^2}{9} &= 1 - \frac{9}{25} = \frac{16}{25}\\ y^2 &= \frac{144}{25}\\ y &= \frac{12}{5} = 2.4. \end{aligned}$$

Check­ing the answer#

$2.4$ m is less than the cen­tral height $3$ m and more than $0$, which makes sense for a point between the cen­tre and the end.

Answer#

The height is $2.4$ m.

Ques­tion 10: Eccen­tric­ity from focal dis­tance and minor axis#

The prob­lem#

The dis­tance between the foci of an ellipse is $8$ and the minor axis is $6$. Find its eccen­tric­ity.

Under­stand­ing the prob­lem#

"Dis­tance between the foci" is $2c$, and "minor axis" is its full length $2b$. You need $\displaystyle e = \frac{c}{a}$, so you must find $a$ first.

The idea#

Halve the given lengths to get $c$ and $b$, then use $a^2 = b^2 + c^2$.

Step-by-step solu­tion#

Step 1. Halve the lengths.

$$2c = 8 \Rightarrow c = 4, \qquad 2b = 6 \Rightarrow b = 3.$$

Step 2. Find $a$.

$$a^2 = b^2 + c^2 = 9 + 16 = 25, \qquad a = 5.$$

Step 3. Find $e$.

$$\displaystyle e = \frac{c}{a} = \frac{4}{5}.$$

Check­ing the answer#

$e = 0.8 < 1$, as needed for an ellipse.

Answer#

$\displaystyle e = \frac{4}{5}$

Com­mon mis­take to avoid#

Do not use $8$ and $6$ directly. They are the full lengths $2c$ and $2b$; halve them first.