Miscellaneous Exercise answers: Conic Sections

Open NCERT p. 204Checked by computer

1. Put the vertex at the origin with the axis of the reflector along the x-axis, so its section is y² = 4ax.
2. The reflector is 20 cm across, so the rim is 10 cm on each side of the axis; it is 5 cm deep, so the rim point (5, 10) lies on the parabola.
3. Substitute: 10² = 4a × 5, so 100 = 20a and a = 5.
4. The focus of y² = 4ax is (a, 0), so the focus is (5, 0).

AnswerThe focus is at (5, 0), on the axis 5 cm from the vertex.

Watch this explained “Where it earns its keep”, 10:23 into The chord through the focus that measures how open the curve is

Open NCERT p. 204Checked by computer

1. Put the vertex at the top, with x measured across and y measured downward from the vertex: x² = 4ay.
2. At the base, y = 10 m and the half-width is 2.5 m (half of 5 m).
3. Substitute: (2.5)² = 4a × 10, so a = 5/32.
4. At 2 m from the vertex, y = 2, so (half-width)² = 4a × 2 = 5/4.
5. Half-width = √5/2, so the full width = √5.
6. The answer key at the back of the book prints 2.23 m; the exact width is √5 m = 2.236… m, which rounds to 2.24 m, so the key has cut off the digits instead of rounding.

Answer√5 m (about 2.24 m)

Watch this explained “Three uses, and one honest limit”, 14:16 into Balancing a point against a line, and the four equations that result

Open NCERT p. 204Matches NCERT’s answer

1. Put the lowest point of the cable at the origin, with x measured from the middle and y the height of the cable above that lowest point: x² = 4ay.
2. At the end of the roadway, x = 50 m (half of 100 m); the wire there is 30 m and the shortest (middle) wire is 6 m, so y = 30 − 6 = 24 m.
3. Substitute: 50² = 4a × 24, so a = 625/24.
4. At x = 18 m: y = 18²/(4a) = 1944/625.
5. Wire length = y + 6 (add back the shortest wire) = 1944/625 + 6 = 5694/625.

Answer5694/625 m ≈ 9.11 m

Watch this explained “Three uses, and one honest limit”, 14:16 into Balancing a point against a line, and the four equations that result

Open NCERT p. 204Matches NCERT’s answer

1. Put the centre of the base at the origin: x²/16 + y²/4 = 1 (half-width 4 m, height 2 m), taking y ≥ 0.
2. A point 1.5 m from one end is 4 − 1.5 = 2.5 m from the centre, so x = 2.5.
3. Substitute: (2.5)²/16 + y²/4 = 1, so y² = 39/16.
4. y = √39/4.

Answer√39/4 m (about 1.56 m)

Watch this explained “A rod and an arch”, 14:26 into Putting the centre at the origin, and reading the axes off the equation

Open NCERT p. 204Matches NCERT’s answer

1. Let the rod AB make an angle θ with the x-axis, with A on the x-axis and B on the y-axis, and let P(x, y) be the point with AP = 3 cm.
2. Then PB = 12 − 3 = 9 cm.
3. Draw PQ perpendicular to the y-axis and PR perpendicular to the x-axis. Then PQ = x and PR = y.
4. In triangle PBQ the angle at P is θ, so cos θ = PQ/PB = x/9.
5. In triangle PRA the angle at A is θ, so sin θ = PR/PA = y/3.
6. Since cos²θ + sin²θ = 1: (x/9)² + (y/3)² = 1, that is x²/81 + y²/9 = 1.

Answerx²/81 + y²/9 = 1

Watch this explained “A rod and an arch”, 14:26 into Putting the centre at the origin, and reading the axes off the equation

Open NCERT p. 204Matches NCERT’s answer

1. Compare x² = 12y with x² = 4ay: 4a = 12, so a = 3.
2. The ends of the latus rectum are (2a, a) and (−2a, a), i.e. (6, 3) and (−6, 3).
3. The triangle has vertex (0, 0) and base from (−6, 3) to (6, 3), so base = 12 and height = 3.
4. Area = ½ × base × height = ½ × 12 × 3 = 18.

Answer18 square units

Watch this explained “Where it earns its keep”, 10:23 into The chord through the focus that measures how open the curve is

Open NCERT p. 204Matches NCERT’s answer

1. The locus of points where the sum of distances from two fixed points is constant is an ellipse.
2. Given: sum of distances = 10 m, so 2a = 10, hence a = 5.
3. Distance between flag posts = 2c = 8 m, so c = 4.
4. Use the relationship b² = a² − c² = 25 − 16 = 9, so b = 3.
5. Place flag posts on the x-axis at (±4, 0).
6. Standard ellipse equation: x²/a² + y²/b² = 1.
7. Substitute a = 5 and b = 3: x²/25 + y²/9 = 1.

Answerx²/25 + y²/9 = 1

Watch this explained “Check before you compute”, 10:39 into Two fixed points and a fixed total distance

Open NCERT p. 204Matches NCERT’s answer

1. Vertex of parabola at (0, 0).
2. Other two vertices by symmetry: (x₁, y₁) and (x₁, −y₁).
3. For equilateral triangle, all sides equal.
4. Distance from (0, 0) to (x₁, y₁): √(x₁² + y₁²).
5. Distance from (x₁, y₁) to (x₁, −y₁): 2y₁.
6. Equate: √(x₁² + y₁²) = 2y₁.
7. Square: x₁² + y₁² = 4y₁².
8. Simplify: x₁² = 3y₁².
9. Since (x₁, y₁) is on parabola: y₁² = 4ax₁.
10. Substitute: x₁² = 3(4ax₁) = 12ax₁.
11. Solve: x₁ = 12a.
12. Then: y₁² = 4a(12a) = 48a², so y₁ = 4a√3.
13. Side length = 2y₁ = 8a√3.

Answer8a√3

Watch the lesson Balancing a point against a line, and the four equations that result