Learn Analytic Geometry Online

Another name of Analytical  geometry is co-ordinate geometry and it describes as a graph of quadratic equations in the co-ordinate plane. Analytical geometry grew out of need for establishing uniform techniques for solving geometrical problems, the aim being to apply them to study of curves, which are of particular importance in practical problems. In this article we shall discuss the learning analytic geometry ans some example problems

Example Problem 1 to learn analytic geometry online:

Find the equation of the parabola if the curve is open upward, vertex is (− 1, − 2) and the length of the latus rectum is 4.

Solution:

Since it is open upward, the equation is of the form

(x − h)2 = 4a(y − k)

Length of the latus rectum = 4a = 4 and this gives a = 1

The vertex V (h, k) is (− 1, − 2)

The equation of parabola is [x-(-1)]2=4*1 [y-(-2)]

[x+1]2=4[y+2]

Example Problem 2 - Learn analytic geometry online  :

Find the equation of the parabola if

(i) the vertex is (0, 0) and the focus is (− a, 0), a > 0

Solution: (i) From the given data of the parabola is open leftward

The equation of the parabola is of the form

(y − k)2 = − 4a(x − h)

Here, the vertex (h, k) is (0, 0) and VF = a

The required equation is

(y − 0)2 = − 4a (x − 0)

y2 = − 4ax.

Example Problem 3 - learn analytic geometry online :

Find the equation of the parabola if the curve is open rightward, vertex is (2, 1) and passing through point (6,5).

Solution: Since it is open rightward, the equation of the parabola is of the form

(y − k)2 = 4a(x − h)

The vertex V(h, k) is (2, 1)

∴ (y − 1)2 = 4a (x − 2)

But it passes through (6, 5)

(5-1)2 = 4a (6 − 2) -- > 16= 4a * 16

4a= 1 ---- > a = 1/4

The required equation is (y − 1)2 = 1/4 (x − 2)

Example Problem 4 - learn analytic geometry online :

Find the equation of the parabola if the curve is open leftward, vertex is (2, 0) and the distance between the latus rectum and directrix is 2.

Solution: Since it is open leftward, the equation is of the form

(y − k)2 = − 4a(x − h)

The vertex V(h, k) is (2, 0)

The distance between latus rectum and directrix = 2a = 2 giving a = 1 and the equation of the parabola is

(y − 0)2 = − 4(1) (x − 2)

or y2 = − 4(x − 2)

Practice problems- learn analytic geometry online:

1.Find the equation of the parabola whose vertex are (1, 2) and the equation of the directrix x = 3.

The required equation is

(y − 2)2 = 4(2) (x − 1)

(y − 2)2 = 8(x − 1)

2.The separate equations of the asymptotes of the hyperbola 4x2-25y2=100

Answer   :x/5-y/2=0 and x/5+y/2=0