# Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.3

## 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:
Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A = 2, C = – 1
Elere A ≠ C also A and C are of opposite signs.
So the conic 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:
Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A = C and B = 0
So the given conic 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 of opposite sign.
So the conic is a hyperbola.

Question 6.
y2 + 4x + 3y + 4 = 0
Solution:
Comparing this equation with the general equation of the conic
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
We get A = 0 and B = 0
So the conic 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
(i) x2 – 4y2 + 6x + 16y – 11 = 0
(ii) y2 – 8y + 4x – 3 = 0
(iii) 4x2 – 9y2 = 36
(iv) 16x2 + 25y2 = 400
(v) 16x2 + 9y2 + 32x – 36y – 92 = 0
(vi) x2 + 4y2 – 8x – 16y – 68 = 0
(vii) x2 + y2 – 4x + 6y – 17 = 0
Solution:
(i) Hyperbola
(ii) Parabola
(iii) Hyperbola
(iv) Ellipse
(v) Ellipse
(vi) Ellipse
(vii) Circle