ilovepdf_merged (6).pdf). Content covers sections 16.1–16.x.Conics arise as plane sections of a cone. NIOS focuses on standard equations of the parabola, ellipse and hyperbola, with focus–directrix definitions and basic properties.
Definition: set of points equidistant from a fixed point (focus) and a fixed line (directrix).
Standard form x²/a² + y²/b² = 1 with a > b > 0: major axis along x, foci (±ae, 0) where e = √(1 − b²/a²) and 0 < e < 1. Vertices (±a,0); co-vertices (0,±b).
Standard x²/a² − y²/b² = 1: foci (±ae,0) with e = √(1 + b²/a²) > 1. Asymptotes y = ±(b/a)x.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Conic Sections.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: distance from a fixed point is always in a constant ratio to its perpendicular distance
Memorise and apply: The constant ratio is called the eccentricity and is denoted by e.
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style identify conic, Oct 2024, Sample / board ellipse, Sample QP 2024 (conic standard)
PYQ1. For the hyperbola x²/16 − y²/9 = 1, the length of the latus rectum is:
Model Answer
1) 9/2 units
Explanation
1) a²=16, b²=
9. Latus rectum of hyperbola = 2b²/a = 2·9/4 = 9/2.
PYQ2. For the parabola y² = 16x, find the focus and the equation of the directrix.
Model Answer
1) Focus (4, 0); directrix x = −4
Explanation
1) y²=4ax with 4a=16 ⇒ a=
4. Focus (a,0)=(4,0); directrix x=−a=−4.
PYQ3. For the ellipse x²/25 + y²/9 = 1, find the eccentricity e and the foci.
Model Answer
1) e = 4/5; foci (±4, 0)
Explanation
1) a=5, b=
3. e=√(1−b²/a²)=√(1−9/25)=√(16/25)=4/
5. Foci (±ae,0)=(±4,0).
PYQ4. Identify the conic represented by 9x² + 4y² = 36.
Model Answer
1) Ellipse: x²/4 + y²/9 = 1
Explanation
1) Divide by 36: x²/4 + y²/9 = 1 — sum of positive square terms equals 1
2) ellipse.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
For the parabola y² = 16x, find the focus and the equation of the directrix.
1) Compare y²=16x with y²=4ax ⇒ 4a=16 ⇒ a=4.
2) Focus = (a, 0) = (4, 0).
3) Directrix: x = −a ⇒ x = −4.
Answer: Focus (4, 0); directrix x = −4
Standard parabola y²=4ax has focus (a,0) and directrix x=−a.
4a=16 so a=4; focus four units right of origin; directrix four units left.
Vertex is (0,0); axis is the x-axis.
Identify a from 4a, not from the coefficient alone without dividing by 4.
For the ellipse x²/25 + y²/9 = 1, find a, b and the eccentricity e.
1) a²=25 ⇒ a=5; b²=9 ⇒ b=3 (a>b).
2) e = √(1 − b²/a²) = √(1 − 9/25) = √(16/25) = 4/5.
Answer: a=5, b=3, e=4/5
From the standard ellipse with major axis along x, eccentricity is √(1−b²/a²)=4/5.
Semi-major 5, semi-minor 3; e=√(1−9/25)=4/5.
Foci are (±ae,0)=(±4,0).
Ensure a>b before using e=√(1−b²/a²).
For x²/16 − y²/9 = 1, find e and the foci.
1) a²=16 ⇒ a=4; b²=9 ⇒ b=3.
2) e = √(1 + 9/16) = √(25/16) = 5/4.
3) Foci (±ae, 0) = (±5, 0).
Answer: e=5/4; foci (±5, 0)
Hyperbola eccentricity √(1+b²/a²) and foci (±ae,0).
a=4,b=3 → e=5/4 → foci at x=±5.
Asymptotes y=±(b/a)x = ±(3/4)x.
Hyperbola uses + inside the square root for e.
Find the length of the latus rectum of y² = 12x.
1) 4a = 12 ⇒ a = 3.
2) Length of latus rectum = 4a = 12.
Answer: 12
For y²=4ax, latus rectum has length 4a; here 4a=12.
Coefficient 12 is already 4a, so latus rectum length is 12.
Latus rectum is the focal chord parallel to the directrix.
Do not report a=12.
Identify the conic: 9x² + 4y² = 36.
1) Divide by 36: x²/4 + y²/9 = 1.
2) This is an ellipse (sum of positive square terms = 1).
3) Here b²=9>a²=4 so major axis is along y (semi-axes 2 and 3).
Answer: Ellipse: x²/4 + y²/9 = 1
Reducing to standard form shows an ellipse.
Both terms positive with =1 after dividing → ellipse.
If it were minus between terms → hyperbola.
Always divide to make RHS 1 when possible.
Find the foci of x²/169 + y²/144 = 1.
1) a²=169 ⇒ a=13; b²=144 ⇒ b=12.
2) e = √(1−144/169)=√(25/169)=5/13.
3) ae = 13·5/13 = 5 ⇒ foci (±5, 0).
Answer: Foci (±5, 0)
ae=5 gives foci on the major axis at (±5,0).
a=13,b=12,e=5/13, so ae=5.
c=ae is sometimes used for linear eccentricity.
a must be the semi-major axis (larger denominator under the axis variable).
For y² = 8x, find a and the focus.
1) 4a = 8 ⇒ a = 2.
2) Focus (a,0) = (2, 0).
Answer: a=2; focus (2, 0)
Compare with y²=4ax.
Focus is a units to the right of the vertex.
Directrix would be x=−2.
Check each algebraic step carefully.
Find the eccentricity of x²/100 + y²/36 = 1.
1) a=10, b=6.
2) e = √(1 − 36/100) = √(64/100) = 8/10 = 4/5.
Answer: e = 4/5
Standard ellipse formula with a>b.
1 − 0.36 = 0.64; √0.64 = 0.8.
Foci at (±ae,0)=(±8,0).
Check each algebraic step carefully.
Write the asymptotes of x²/9 − y²/4 = 1.
1) a=3, b=2.
2) Asymptotes: y = ±(2/3)x.
Answer: y = ±(2/3)x
Asymptotes of the standard hyperbola x²/a²−y²/b²=1.
Slope is ±b/a.
They pass through the origin.
Check each algebraic step carefully.
Identify: 16x² − 9y² = 144.
1) Divide by 144: x²/9 − y²/16 = 1.
2) Difference of squares = 1 ⇒ hyperbola.
Answer: Hyperbola: x²/9 − y²/16 = 1
Reduce to standard form to identify.
Minus between terms means hyperbola.
a²=9, b²=16.
Check each algebraic step carefully.