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# Analytical Geometry

Analytical Geometry explains the concepts of coordinate system, idea of locus of a point. They are mainly used to understand the XY plane. In such plane, we deal with the conic sections too.

We deal with three conic sections like parabola, ellipse and hyperbola, which are explained in a brief manner along with problems. Analytical geometry is also referred as Cartesian geometry. Let us discuss about few of curves drawn in a coordinate system.

The intersection of the plane with the cone is either at the vertex of the cone or any other part:

Parabola: Parabola is the locus of a point, which moves on a plane and its distance from the focus is equal to the distance from directrix.

Equation of parabola in standard form: y$^2$ = 4ax

Ellipse: Ellipse is the locus of a point, which moves in such a way that the ratio of its distance from the focus to its distance from directrix is less than 1.

Equation of Ellipse in standard form:  $\frac{x^{2}}{a^{2}}$$\frac{y^{2}}{b^{2}}$ = 1

Hyperbola: Hyperbola is the locus of a point, which moves in such a way that the ratio of its distance from the focus to its distance from directrix is greater than 1.

Equation of Hyperbola in standard form:  $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$=1

Let us work out some sample problems of analytical geometry.

 Related Calculators Calculate Geometry

## What is Analytical Geometry

Analytical geometry is also known as analytic geometry. It was one of the biggest inventions which is said to be done by Descartes and Fermat. Analytic geometry is a branch in mathematics which models geometric shapes, such as points, lines, circles and other shapes. In fact in mathematics, the analytical geometry is also termed as the Cartesian geometry or the coordinate geometry. It is said to be the study concerned of geometry which uses a coordinate system.

Except mathematics, the analytical geometry is commonly seen in engineering as well as in physics. It serves as the foundation for most of the mathematical areas, such as - algebra, geometry, calculus, computational geometry etc.

## Problems Based on Points and Lines

In this section, we shall discuss few problems related to points and lines. Have a look.
Example 1 : Find the measure of length of a line segment whose endpoints are (-3, 1) and (-6, -5).
Solution : This length can be calculated using the distance formula which measures the distance between two points located in a plane. According to distance formula,
d = $\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$
Here, $x_{1}$ = -3, $y_{1}$ =  1, $x_{2}$ = -6, $y_{2}$ = -5
Plugging values in above formula, we get
d = $\sqrt{(-3 - (-6))^{2} + (1 - (-5))^{2}}$
= $\sqrt{(-3 + 6)^{2} + (1 + 5)^{2}}$
= $\sqrt{3^{2} + 6^{2}}$
= $\sqrt{9 + 36}$
= $\sqrt{45}$ = 9$\sqrt{5}$ units

Example 2 : In a math class, three columns of seats were located. Betty's seat in the left most corner is located at the point (-4, 7), while Mary's seat at the right most corner is situated at the point (-2, 5). What would be the coordinates of the middle seat at the line joining Betty's and Mary's seat ?
Solution : Here, we shall use midpoint formula which is,
x = $\frac{x_{1}+ x_{2}}{2}$ and y = $\frac{y_{1}+ y_{2}}{2}$
where, (x, y) is denoted by the midpoint of a line segment whose end points are ($x_{1}, y_{1}$) and ($x_{2}, y_{2}$).
x = $\frac{x_{1}+ x_{2}}{2}$ and y = $\frac{y_{1}+ y_{2}}{2}$

Here, $x_{1}$ = -4, $x_{2}$ = -2 and $y_{1}$ = 7, $y_{2}$ = 5
Substituting above values in the formulas.
x = $\frac{-4+ (-2)}{2}$ and y = $\frac{7+ 5}{2}$
x = $\frac{-6}{2}$ and y = $\frac{12}{2}$
x = -3 and y = 6
Co-ordinates of the middle seat are (-3, 6).

Example 3 : The two parallel lines are given by the equations 4x + 3y = 5 and 4x + 3y = 10. Determine the distance between them.
Solution : The perpendicular distance between two parallel lines is constant. Thus, we can calculate it using the following formula.
Suppose two parallel lines are gives by :
ax + by + c$_{1}$ = 0
and
ax + by + c$_{2}$ = 0

Then, distance (d) between them is determined by :
d = $\frac{|c_{1}-c_{2}|}{\sqrt{a^{2}+b^{2}}}$

The given equations are :
4x + 3y = 5 and 4x + 3y = 10
or
4x + 3y - 5 = 0 and 4x + 3y - 10 = 0

Here, a = 4, b = 3, c$_{1}$ = - 5 and c$_{2}$ = -10
Plugging these values in above formula :
d = $\frac{|c_{1}-c_{2}|}{\sqrt{a^{2}+b^{2}}}$
= $\frac{|-5-(-10)|}{\sqrt{4^{2}+3^{2}}}$
= $\frac{5}{\sqrt{16+9}}$
= $\frac{5}{\sqrt{25}}$
= $\frac{5}{5}$
= 1 unit.

## Problems Based on Parabola

From the given image, Focus (F) is the fixed point, Directrix (t) is a fixed line. The distance between the given focus and directrix is called focal parameter (p).

Coordinates of the focus: F($\frac{p}{2}$, 0)

Coordinates of the vertex: M(0, 0)

Example 1: 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)

$4^2$ = 4a (4)

16 = 16a

Divide by 16 on both sides

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

a = 1

$(y - 1)^2$ = 4(x - 2)

Example 2: Find the co-ordinate of end point of the latus rectum of the parabola $y^2$ = 12x

Solution: Standard form of Parabola $y^2$ = 4ax

Compare given equation and standard equation, we get
4a = 12

Divide by constant term 4 on both side of equation

a = 3

Co-ordinate of endpoints of latus rectum of parabola (a, 2a) and (a, -2a)

= (3, 6) and (3, -6)

## Problems Based on Ellipse

Ellipse is a closed curve in nature. From the following figure, the distance from focus to the center is called focal distance. S is the focus and l is the directrix.

Example 1: 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

$\frac{(x - h)^2}{b^2}$+$\frac{(y - k)^2}{a^2}$ = 1

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

Semi major axis a = 12, $a^2$ = 144

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

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

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

$b^2$ = 36

b = 6
$\frac{(x - 1)^2}{36}$+$\frac{(y - 12)^2}{144}$ = 1

## Problems Based on Hyperbola

Hyperbola is a smooth curve, which consists of two distinct and similar branches of curve. From the figure distance between two foci  (F1, F2) is called as focal distance.

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

Solution:

From the given data, the equation is in the form

$\frac{(x - h)^2}{a^2}$ - $\frac{(y - k)^2}{b^2}$ = 1

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

Where ae = 8 - 2 = 6 ___(i)

The distance between the centre and directrix $\frac{a}{e}$ = 2

So ae $\times$ $\frac{a}{e}$ = 6 * 2 = 12
$e^2$= 3

Squaring equation (i) and plugging the value of a$^{ 2}$, we get

$a^{2}e^{2}$ = 36
$a^{2}$ = 12

We have
$b^2$= $a^2(e^2 - 1)$
= 12(3 - 1)
= 12 $\times$ 2
= 24

$\frac{(x-2)^2}{12}$- $\frac{(y-1)^2}{24}$ = 1

 More topics in Analytical Geometry Cartesian System Parametric Equation of a Circle Asymptote Distance Time Equation Parabola Equation of Curves Conic Sections Hyperbola Ellipse Latus Rectum Jacobian Line of Intersection of planes
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