Conic Sections · Foci & Eccentricity
Morph between an ellipse, hyperbola and parabola; study eccentricity, foci, directrices and standard equations on a dynamic coordinate plane.
NCERT alignment: Class 11; chapter Conic Sections. Exam practice: JEE Main, JEE Advanced.
Learning objectives
Students should use the live simulator to test a quantitative prediction, read measurements and explain the limits of the model. Do not rely on the animation alone: calculate the expected result before adjusting controls.
Theory, equations and interpretations
Ellipse locus
For a horizontal major axis (a>b), the sum of distances from the two foci is 2a. The foci are (h±c,k), where c²=a²−b².
Important relation: (x−h)²/a² + (y−k)²/b² = 1
Hyperbola locus
The absolute difference of focal distances equals 2a. Here c²=a²+b² and eccentricity e=c/a>1.
Important relation: (x−h)²/a² − (y−k)²/b² = 1
Parabola as focus-directrix locus
Each point has equal distance from focus (h+p,k) and directrix x=h−p. The vertex is (h,k) and e=1.
Important relation: (y−k)² = 4p(x−h)
Eccentricity classification
Eccentricity measures the shape. Ratios and axis conditions in the current model are explicit so no impossible ellipse is drawn.
Important relation: Ellipse: 0<e<1; Parabola: e=1; Hyperbola: e>1
Practical investigations
- Ellipse foci: Fix a=2 and lower b/a from 0.9 to 0.4. Predict how far the foci move from the center.
- Hyperbola asymptotes: With a=2 and b/a=0.5, inspect the dashed asymptotes y−k=±(b/a)(x−h).
- Parabolic locus: Switch to parabola. Move the probe and compare focus distance to directrix distance.
- Translate without deforming: Change h,k without changing a or b/a. What remains unchanged?
Worked exam-style questions and answers
Q1. Ellipse x²/25+y²/9=1 has c=
Options: A. 3; B. 4; C. 5; D. √34
Answer: B. c²=a²−b²=25−9=16, so c=4.
Q2. For a standard ellipse, eccentricity is
Options: A. 0<e<1; B. e=1; C. e>1; D. e<0
Answer: A. Ellipse e=c/a with 0<c<a.
Q3. For y²=8x, focus is
Options: A. (0,2); B. (2,0); C. (−2,0); D. (8,0)
Answer: B. 4p=8 gives p=2 and focus (2,0).
Q4. Hyperbola x²/9−y²/16=1 has c=
Options: A. 4; B. 3; C. 5; D. 7
Answer: C. c²=9+16=25 gives c=5.
Q5. Parabola has eccentricity
Options: A. 0; B. 1; C. 2; D. variable >2
Answer: B. Distance to focus equals distance to directrix, so e=1.
Q6. Asymptote slopes of x²/a²−y²/b²=1 are
Options: A. ±a/b; B. ±b/a; C. ±1; D. 0
Answer: B. Asymptotes are y=±(b/a)x.
Common misconceptions and limitations
Axes are aligned with coordinates; no general rotation is modeled. For ellipse b=ra with 0<r<1; for hyperbola b=ra; parabola uses y²=4px with p>0. Directrices for conics can extend beyond the viewport.
All learning content is available free of charge. The simulation is a simplified educational model. Check assumptions and units before applying a result. NANDAX SimStudio is not affiliated with NCERT or examination boards.
Teacher classroom use
Ask students to form a hypothesis, vary one parameter at a time, keep a record of graph readings, explain any change and compare results with the analytic relation. The on-screen worksheet offers questions with an answer key.