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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.3
Identify the type of conic section for each of the equations.
Question 1.
2x2 – y2 = 7
Solution:
A = 2, B = 0, C = -1, F = -7
A ≠ C, A, and C are of opposite signs.
∴ It is a hyperbola.
Question 2.
3x2 + 3y2 – 4x + 3y + 10 = 0
Sol. Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A = C also B = 0
So the given conic is a circle.
Question 3.
3x2 + 2y2 = 14
Solution:
Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A ≠ C also C are of the same sign.
So the given conic is an ellipse.
Question 4.
x2 + y2 + x – y = 0
Solution:
A = 1, B = 0, C = 1, D = 1, E = -1
A = C and B = 0 (No xy term)
∴ It is a circle.
Question 5.
11x2 – 25y2 – 44x + 50y – 256 = 0
Solution:
Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A ≠ C. Also A and C are opposite signs.
So the conic is a hyperbola.
Question 6.
y2 + 4x + 3y + 4 = 0
Solution:
A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0 and A = 0 (either A or C = 0)
∴ It is a parabola.
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.3 Additional Problems
Identify the type of conic section for each of the following equations
- x2 – 4y2 + 6x + 16y – 11 = 0
- y2 – 8y + 4x – 3 = 0
- 4x2 – 9y2 = 36
- 16x2 + 25y2 = 400
- 16x2 + 9y2 + 32x – 36y – 92 = 0
- x2 + 4y2 – 8x – 16y – 68 = 0
- x2 + y2 – 4x + 6y – 17 = 0
Solution:
- Hyperbola
- Parabola
- Hyperbola
- Ellipse
- Ellipse
- Ellipse
- Circle