Solved Examples

Solved Examples — the Full Chapter Workout

Thirty worked problems: circle synthesis, the full applied set (mirrors, beams, bridges, arches, rods, racecourses), cross-conic drills and eccentricity identities. Attempt each before reading the solution.

Example 1: The parabolic mirror

The focus of a parabolic mirror is 5 cm from its vertex. If the mirror is 45 cm deep, find the distance $AB$ across the open end.

Solution:

Step 1 — Model. Vertex at the origin, axis along the positive $x$-axis: $a = 5$, so $y^2 = 20x$.

Step 2 — Evaluate at the rim. Depth $x = 45$: $y^2 = 900$, so $y = \pm 30$.

Step 3 — Read the width. $AB = 30 - (-30) = 60$ cm.

Takeaway: Place the vertex at the origin and the axis along a coordinate axis — the physics then reads straight off the equation.

Example 2: The deflected beam

A beam supported at ends 12 m apart deflects 3 cm at the centre, taking a parabolic shape. How far from the centre is the deflection 1 cm?

Solution:

Step 1 — Model from the lowest point. Vertex at the centre's deflected position, opening up: $x^2 = 4ay$ through $\left(6, \frac{3}{100}\right)$.

Step 2 — Find $4a$. $36 = 4a \cdot \frac{3}{100}$ gives $4a = 1200$.

Step 3 — Deflection 1 cm. That point sits $\frac{2}{100}$ m above the vertex: $x^2 = 1200 \times \frac{2}{100} = 24$: $x = 2\sqrt{6}$ m.

Takeaway: Convert units once at the start (cm → m) and the parabola handles the rest.

Example 3: The sliding rod (15 cm)

A rod $AB$ of length 15 cm slides with $A$ on the $x$-axis and $B$ on the $y$-axis. $P$ on the rod has $AP = 6$ cm. Show that the locus of $P$ is an ellipse.

Solution:

Step 1 — Parametrise by the rod's angle. With the rod at angle $\theta$ to the $x$-axis and $PB = 15 - 6 = 9$: dropping perpendiculars, $x = 9\cos\theta$ and $y = 6\sin\theta$.

Step 2 — Eliminate $\theta$. $\cos^2\theta + \sin^2\theta = 1$: $\frac{x^2}{81} + \frac{y^2}{36} = 1$.

Step 3 — Conclude. The locus is an ellipse with $a = 9$, $b = 6$. ∎

Takeaway: $\cos^2 + \sin^2 = 1$ is the standard eliminator — express $x$ and $y$ in terms of one angle and square-add.

Example 4: A concentric circle

Find the circle concentric with $x^2 + y^2 - 4x - 8y - 45 = 0$ and passing through the origin.

Solution:

Step 1 — Extract the centre. Completing squares: centre $(2, 4)$.

Step 2 — New radius. Through $(0, 0)$: $r^2 = 4 + 16 = 20$.

Step 3 — Write. $(x - 2)^2 + (y - 4)^2 = 20$, i.e. $x^2 + y^2 - 4x - 8y = 0$.

Takeaway: Concentric = same centre, new radius — only the constant term changes in the expanded form.

Example 5: Circle from the ends of a diameter

Find the circle with $(1, 2)$ and $(3, -4)$ as ends of a diameter.

Solution:

Step 1 — Centre = midpoint. $\left(\frac{1+3}{2}, \frac{2-4}{2}\right) = (2, -1)$.

Step 2 — Radius from either end. $r^2 = (1-2)^2 + (2+1)^2 = 1 + 9 = 10$.

Step 3 — Write. $(x - 2)^2 + (y + 1)^2 = 10$.

Takeaway: Diameter ends give the centre for free — midpoint first, then one distance.

Example 6: Anatomy plus a point test

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$, and check whether $(0, 2)$ lies on it.

Solution:

Step 1 — Complete squares. $(x - 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25$: centre $(3, -2)$, radius 5.

Step 2 — Test the point. Distance$^2$ from $(3, -2)$ to $(0, 2)$: $9 + 16 = 25$.

Step 3 — Conclude. Equal to $r^2$: the point lies ON the circle.

Takeaway: The on/inside/outside test is one squared-distance comparison — no roots required.

Example 7: The reflector's focus

A parabolic reflector is 20 cm in diameter and 5 cm deep. Find the focus.

Solution:

Step 1 — Model. Vertex at origin, axis along $x$: the rim point $(5, 10)$ lies on $y^2 = 4ax$.

Step 2 — Solve for $a$. $100 = 20a$: $a = 5$.

Step 3 — Conclude. Focus $(5, 0)$ — 5 cm from the vertex, exactly in the plane of the rim.

Takeaway: A reflector focused at its rim plane is a designed coincidence here — the numbers $(5, 10)$ force it.

Example 8: The parabolic arch

An arch shaped like a parabola with vertical axis is 10 m high and 5 m wide at the base. How wide is it 2 m below the vertex?

Solution:

Step 1 — Model downward from the vertex. Measuring $y$ downward from the top: $x^2 = 4ay$ through $(2.5, 10)$.

Step 2 — Find $4a$. $6.25 = 4a(10)$: $4a = \frac{5}{8}$.

Step 3 — Evaluate at $y = 2$. $x^2 = \frac{5}{8} \times 2 = \frac{5}{4}$: $x = \frac{\sqrt{5}}{2}$, so the width is $2x = \sqrt{5} \approx 2.23$ m.

Takeaway: Choosing the measuring direction (downward) keeps every number positive — a free simplification.

Example 9: The suspension bridge

A suspension-bridge cable hangs as a parabola over a 100 m roadway; the longest vertical wire is 30 m and the shortest (central) is 6 m. Find the wire 18 m from the middle.

Solution:

Step 1 — Model from the cable's lowest point. Vertex there; the cable rises $30 - 6 = 24$ m over the half-span 50 m: $x^2 = 4ay$ through $(50, 24)$.

Step 2 — Find $4a$. $2500 = 4a(24)$: $4a = \frac{2500}{24} = \frac{625}{6}$.

Step 3 — Sag at $x = 18$. $y = \frac{x^2}{4a} = \frac{324 \times 24}{2500} = \frac{7776}{2500} \approx 3.11$ m.

Step 4 — Wire length. $6 + 3.11 = 9.11$ m (approx).

Takeaway: Wires measure roadway-to-cable: add the central clearance back after computing the parabolic rise.

Example 10: Triangle on the latus rectum

Find the area of the triangle formed by joining the vertex of $x^2 = 12y$ to the ends of its latus rectum.

Solution:

Step 1 — Locate the latus rectum. $4a = 12$: $a = 3$; ends $(\pm 6, 3)$.

Step 2 — Base and height. Base $= 12$ (horizontal chord); height $= 3$ (vertex to chord).

Step 3 — Area. $\frac{1}{2} \times 12 \times 3 = 18$.

Takeaway: The latus rectum triangle of $x^2 = 4ay$ always has area $\frac{1}{2}(4a)(a) = 2a^2$ — here $2(9) = 18$ ✓.

Example 11: The inscribed equilateral triangle

An equilateral triangle is inscribed in $y^2 = 4ax$ with one vertex at the parabola's vertex. Find its side.

Solution:

Step 1 — Set up by symmetry. The other two vertices are $(h, \pm k)$; the side from the origin makes $30°$ with the axis: $\frac{k}{h} = \tan 30° = \frac{1}{\sqrt{3}}$.

Step 2 — Use the parabola. $k^2 = 4ah$ with $h = k\sqrt{3}$: $k^2 = 4ak\sqrt{3}$, so $k = 4\sqrt{3}a$ (and $h = 12a$).

Step 3 — Side length. The vertical side is $2k = 8\sqrt{3}a$.

Takeaway: Symmetry about the axis halves the work — impose the half-angle $30°$, not $60°$.

Example 12: A chord of $x^2 = 6y$

Find the points of $x^2 = 6y$ at height $y = 6$, and the chord length between them.

Solution:

Step 1 — Substitute the height. $x^2 = 36$: $x = \pm 6$.

Step 2 — Chord. From $(-6, 6)$ to $(6, 6)$: length 12.

Takeaway: Horizontal chords of $x^2 = 4ay$ have length $2\sqrt{4ay}$ — direct substitution, no formulas.

Example 13: The semi-elliptic arch

An arch is a semi-ellipse, 8 m wide and 2 m high. Find its height 1.5 m from one end.

Solution:

Step 1 — Model. Centre at the base's midpoint: $\frac{x^2}{16} + \frac{y^2}{4} = 1$ ($a = 4$, $b = 2$).

Step 2 — Convert the position. 1.5 m from the end means $x = 4 - 1.5 = 2.5$.

Step 3 — Solve for the height. $y^2 = 4\left(1 - \frac{6.25}{16}\right) = 4 \cdot \frac{9.75}{16} = \frac{39}{16}$: height $= \frac{\sqrt{39}}{4} \approx 1.56$ m.

Takeaway: "Distance from the end" must be converted to a coordinate from the CENTRE before substituting.

Example 14: The sliding rod (12 cm)

A 12 cm rod slides with its ends on the axes. Find the locus of the point $P$ 3 cm from the end touching the $x$-axis.

Solution:

Step 1 — Parametrise. $AP = 3$, $PB = 9$: with rod angle $\theta$, $x = 9\cos\theta$, $y = 3\sin\theta$.

Step 2 — Eliminate. $\frac{x^2}{81} + \frac{y^2}{9} = 1$.

Takeaway: The distances to the two rod-ends become the two semi-axes — $PB$ horizontal, $AP$ vertical.

Example 15: The racecourse ellipse

A runner keeps the sum of distances from two flag posts equal to 10 m; the posts are 8 m apart. Find the path.

Solution:

Step 1 — Translate to ellipse data. Constant sum $2a = 10$; focal separation $2c = 8$: $a = 5$, $c = 4$.

Step 2 — Find $b^2$. $b^2 = 25 - 16 = 9$.

Step 3 — Write. $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Takeaway: The definition IS the model — flag posts are foci, the rope length is $2a$.

Example 16: Ellipse anatomy — $\frac{x^2}{49} + \frac{y^2}{36} = 1$

Find the foci, eccentricity and latus rectum.

Solution:

Step 1 — Read. $a = 7$, $b = 6$, major on $x$.

Step 2 — Compute. $c = \sqrt{49 - 36} = \sqrt{13}$.

Step 3 — List. Foci $(\pm\sqrt{13}, 0)$, $e = \frac{\sqrt{13}}{7}$, latus rectum $\frac{2(36)}{7} = \frac{72}{7}$.

Takeaway: Surd values of $c$ are normal — leave them exact.

Example 17: Ellipse anatomy — $\frac{x^2}{100} + \frac{y^2}{400} = 1$

Find the foci, eccentricity and latus rectum.

Solution:

Step 1 — Orient. $400 > 100$: major on $y$, $a = 20$, $b = 10$.

Step 2 — Compute. $c = \sqrt{400 - 100} = 10\sqrt{3}$.

Step 3 — List. Foci $(0, \pm 10\sqrt{3})$, $e = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$, latus rectum $\frac{2(100)}{20} = 10$.

Takeaway: $e = \frac{\sqrt{3}}{2}$ marks quite a flat ellipse — large $e$ means strong elongation.

Example 18: Building an ellipse from $e$ and the foci

Find the ellipse with foci $(\pm 2, 0)$ and eccentricity $\frac{1}{2}$.

Solution:

Step 1 — Recover $a$. $c = 2$ and $e = \frac{c}{a} = \frac{1}{2}$: $a = 4$.

Step 2 — Find $b^2$. $b^2 = 16 - 4 = 12$.

Step 3 — Write. $\frac{x^2}{16} + \frac{y^2}{12} = 1$.

Takeaway: $e$ plus either $a$ or $c$ recovers the other by one division — then the relation gives $b^2$.

Example 19: Hyperbola anatomy — $16x^2 - 9y^2 = 576$

Find the foci, vertices, eccentricity and latus rectum.

Solution:

Step 1 — Normalise. Divide by 576: $\frac{x^2}{36} - \frac{y^2}{64} = 1$: $a = 6$, $b = 8$.

Step 2 — Compute. $c = \sqrt{36 + 64} = 10$.

Step 3 — List. Foci $(\pm 10, 0)$, vertices $(\pm 6, 0)$, $e = \frac{10}{6} = \frac{5}{3}$, latus rectum $\frac{2(64)}{6} = \frac{64}{3}$.

Takeaway: The 6-8-10 triangle hiding inside — hyperbola $a$-$b$-$c$ triples are Pythagorean by construction.

Example 20: Building a hyperbola from $e$ and the foci

Find the hyperbola with foci $(\pm 6, 0)$ and eccentricity 2.

Solution:

Step 1 — Recover $a$. $a = \frac{c}{e} = \frac{6}{2} = 3$.

Step 2 — Find $b^2$. $b^2 = c^2 - a^2 = 36 - 9 = 27$.

Step 3 — Write. $\frac{x^2}{9} - \frac{y^2}{27} = 1$.

Takeaway: Same recovery as the ellipse — only the $b^2$ formula flips to a subtraction from $c^2$.

Example 21: Latus rectum equals transverse axis

For which hyperbolas does the latus rectum equal the transverse axis?

Solution:

Step 1 — Set the condition. $\frac{2b^2}{a} = 2a$.

Step 2 — Simplify. $b^2 = a^2$: $b = a$ — the equilateral hyperbola.

Step 3 — Consequence. $c^2 = 2a^2$: $e = \sqrt{2}$.

Takeaway: Geometric conditions on the latus rectum translate to algebra in $a, b$ in one line.

Example 22: Checking the defining difference

For $\frac{y^2}{25} - \frac{x^2}{39} = 1$, verify at a vertex that the difference of focal distances is $2a$.

Solution:

Step 1 — Anatomy. $a = 5$, $c = \sqrt{25 + 39} = 8$: foci $(0, \pm 8)$, vertex $(0, 5)$.

Step 2 — Two focal distances. From $(0, 5)$: to $(0, 8)$: 3; to $(0, -8)$: 13.

Step 3 — Difference. $13 - 3 = 10 = 2a$ ✓. ∎

Takeaway: Verifying the definition at a vertex takes seconds and confirms $a$ and $c$ simultaneously.

Example 23: Classify the conic

Name the curve: (i) $4x^2 + 9y^2 = 36$; (ii) $4x^2 - 9y^2 = 36$; (iii) $y^2 = 36x$; (iv) $x^2 + y^2 = 36$.

Solution:

Step 1 — (i). Both squares positive, unequal coefficients: ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Step 2 — (ii). Opposite signs: hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$.

Step 3 — (iii). One square only: parabola with $4a = 36$, $a = 9$.

Step 4 — (iv). Equal positive coefficients: circle of radius 6.

Takeaway: Signs and coefficient patterns classify instantly: $++$ unequal = ellipse, $+-$ = hyperbola, one square = parabola, $++$ equal = circle.

Example 24: A quick circle read

Find the centre and radius of $x^2 + y^2 - 2x + 4y - 4 = 0$.

Solution:

Step 1 — Complete squares. $(x - 1)^2 + (y + 2)^2 = 4 + 1 + 4 = 9$.

Step 2 — Read. Centre $(1, -2)$, radius 3.

Takeaway: Half the linear coefficients (sign-flipped) give the centre — a ten-second read once practised.

Example 25: Latus rectum half the minor axis

If the latus rectum of an ellipse equals half its minor axis, find the eccentricity.

Solution:

Step 1 — Set the condition. $\frac{2b^2}{a} = \frac{2b}{2} = b$ gives $2b^2 = ab$, so $a = 2b$.

Step 2 — Convert to $e$. $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}$.

Step 3 — Conclude. $e = \frac{\sqrt{3}}{2}$.

Takeaway: Latus-rectum conditions reduce to the ratio $\frac{b}{a}$, and $e^2 = 1 - \frac{b^2}{a^2}$ finishes.

Example 26: Latus rectum half the major axis

If the latus rectum of an ellipse equals half its major axis, find the eccentricity.

Solution:

Step 1 — Set the condition. $\frac{2b^2}{a} = a$ gives $b^2 = \frac{a^2}{2}$.

Step 2 — Convert. $e^2 = 1 - \frac{1}{2} = \frac{1}{2}$.

Step 3 — Conclude. $e = \frac{1}{\sqrt{2}}$.

Takeaway: Compare with Example 25 — "half the minor" and "half the major" give different, easily-confused answers.

Example 27: A focal-distance preview

The parabola $y^2 = 4ax$ passes through $(3, 6)$. Find $a$, the focus, and the distance from $(3, 6)$ to the focus.

Solution:

Step 1 — Find $a$. $36 = 12a$: $a = 3$; focus $(3, 0)$, directrix $x = -3$.

Step 2 — Focal distance. From $(3, 6)$ to $(3, 0)$: 6.

Step 3 — Confirm via the definition. Directrix distance $= x + a = 3 + 3 = 6$ ✓ — equal, as the parabola demands.

Takeaway: The focal distance of $(x_1, y_1)$ on $y^2 = 4ax$ is always $x_1 + a$ — the definition in formula form.

Example 28: Same numbers, different conics

Compare the distance between the foci of $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Solution:

Step 1 — Ellipse. $c = \sqrt{16 - 9} = \sqrt{7}$: focal distance $2\sqrt{7} \approx 5.29$.

Step 2 — Hyperbola. $c = \sqrt{16 + 9} = 5$: focal distance 10.

Step 3 — Compare. The sign flip pushes the foci from inside the vertices to beyond them.

Takeaway: Identical denominators, different signs: subtract for the ellipse, add for the hyperbola.

Example 29: A parameter from a point

For what $k$ does $x^2 + y^2 - 4x + 6y + k = 0$ pass through $(1, 1)$? Find the radius then.

Solution:

Step 1 — Substitute the point. $1 + 1 - 4 + 6 + k = 0$: $k = -4$.

Step 2 — Complete squares. $(x - 2)^2 + (y + 3)^2 = 4 + 9 - k = 17$.

Step 3 — Read. Radius $\sqrt{17}$.

Takeaway: A point on the curve pins the free parameter first; the geometry follows after.

Example 30: Ellipse from latus rectum and eccentricity

Find the ellipse (major axis on the $x$-axis) with latus rectum 8 and eccentricity $\frac{1}{\sqrt{2}}$.

Solution:

Step 1 — Convert $e$ to a ratio. $b^2 = a^2(1 - e^2) = \frac{a^2}{2}$.

Step 2 — Apply the latus rectum. $\frac{2b^2}{a} = \frac{a^2}{a} = a = 8$.

Step 3 — Write. $a^2 = 64$, $b^2 = 32$: $\frac{x^2}{64} + \frac{y^2}{32} = 1$, foci $(\pm 4\sqrt{2}, 0)$.

Takeaway: Express $b^2$ through $e$ FIRST — the latus rectum then collapses to an equation in $a$ alone.