Analytical Geometry

Learn in this page on the concept of analytical geometry and understand it better with provided examples based on parabola, ellipse and hyperbola.

Analytical geometry is used mainly to understand the plane in xy axis. In such pane we deal with the conic sections. Here , we deal with three conic sections. It is classified according to the position of the intersecting plane. The intersection of the plane with the cone is either at the vertex of the cone or any other part.

i) Analytical geometry problems based on Parabola

ii) Analytical geometry problems based on Ellipse

iii) Analytical geometry problems based on Hyperbola

Let us workout some sample problems in analytical geometry. In the below sections, there are provided workout exampe problems of analytical geometry based on parabola, ellipse and hyperbola.

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

Solution:

Since the curve is open rightward, the equation of a parabola is:

(y - k) ² = 4a(x - h)

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

(y - 1) ² = 4a(x - 2 )

It is passing through the points (6, 5)

So

(5 - 1) ² = 4a (6 - 2)

4² = 4a (4)

16 = 16a

Divide by 16 on both sides

$\frac{16}{16}$ = $\frac{16a}{16}$

a = 1

((y - 1) ² = 4(x - 2)

Find the equation of the ellipse equation of the ellipse if the major axis is parallel to y axis. Semi- major axis is 12, length of the latus rectum is 6 and the center is (1, 12)

Solution:

Since the major axis is parallel to y-axis the equation of the ellipse is of the form

((x - h) ² / b²) + ( (y - k) ² / a²)) = 1

The centre C (h, k) is (1, 12)

Semi major axis a = 12, a² = 144

Length of the latus rectum $\frac{2b^2}{a}$

Where given $\frac{2b^2}{a}$ = 6

$\frac{2b^2}{12}$ = 6

b² = 36

b = 6

((x –1) ² / 36) + ( (y –12) ² / 144)) = 1

Find the equation of a hyperbola equation of a hyperbola whose centre is (2, 1), one of h foci is (8, 1) and the corresponding directrix is x = 4

Solution:

From the given data, the equation is in the form

((x –h) ² / a²) - ( (y –k) ² / b²)) = 1

Where centre (h, k) = (2, 1)

Where ae = 6

The distance between the centre and directrix a / e = 2

So ae x $\frac{a}{e}$ = a² = 6 * 2 = 12

We get e² = 3

b² = a²(e² - 1)

=12(3 - 1)

= 12 * 2

= 24

((x - 2)² / 12) - ( (y - 1)² / 24)) = 1