Parabola equation

X intercept and Y intercept for Parabola equation:
The general form of parabola equation is
y = ax2 + bx + c where a,b and c are parts of the parabola.
Here, x intercepts are the roots of the equation of parabola. The x intercepts are the roots of the equation 0 = ax2 + bx + c. The very common methods to solve the equation are by factoring or by quadratic formula. The y intercept is (0 , c) for parabola equation.

Solve the equations Graphically

Solve the equations Graphically:

Check whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two
solutions of each of the equations, which are given in Table

Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig below We observe that there is a point B (6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, i.e., the given pair of equations is consistent.

General Equation of a Line

General Equation of a Line:
Equation of a Straight line is also called a Linear Equation.

• A straight line is represented by an equation of the first degree in two variables (x and y). Conversely locus of an equation of the first degree in two variables is a straight line.

• A straight line is completely determined by its slope (direction) and a point is given through which the line must pass.

Graph of the equation Ax + By + C = 0 is always a straight line
Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero
simultaneously is called general linear equation or general equation of a line.

Find the area of a square

Problem:
Find the area of the shaded design in below fig where ABCD is a square of side 10 cm a semicircles are drawn with each side of the square as diameter. (Use π = 3.14).


Solution : Let us mark the four unshaded regions as I, II, III and IV .Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm
(10×10-2×1/2×π×52)cm2= (100 – 3.14 × 25) cm2
= (100 – 78.5) cm2 = 21.5 cm2
Similarly, Area of II + Area of IV = 21.5 cm2
So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm2 = (100 – 43) cm2 = 57 cm2

Coordinate plane problem

Problem:
Plot the following ordered pairs of number (x, y) as points in the Cartesian
plane. Use the scale 1cm = 1 unit on the axes.

Solution : The pairs of numbers given in the table can be represented by the points
(– 3, 7), (0, –3.5), (– 1, – 3), (4, 4) and (2, – 3). The locations of the points are shown
by dots in Fig.

Triangle Theorem

Triangle Theorem:
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Proof : Given an arc PQ of a circle subtending angles POQ at the center O and PAQ at a point A on the remaining part of the circle. We need to prove that ∠ POQ = 2 ∠ PAQ.

Consider the three different cases as given in Fig. (i), arc PQ is minor; in (ii),arc PQ is a semicircle and in (iii), arc PQ is major. Let us begin by joining AO and extending it to a point B.
In all the cases, ∠ BOQ = ∠ OAQ + ∠ AQO because an exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Also in Δ OAQ, OA = OQ (Radii of a circle) Therefore, ∠ OAQ = ∠ OQA (Theorem 7.5)
This gives ∠ BOQ = 2 ∠ OAQ (1) Similarly, ∠ BOP = 2 ∠ OAP.... (2) From (1) and (2), ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ) This is the same as ∠ POQ = 2 ∠ PAQ ...(3) For the case (iii), where PQ is the major arc, (3) is replaced by reflex angle POQ = 2 ∠ PAQ

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Construction of Tangents to a Circle

Construction of Tangents to a Circle:

If a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through this point. Therefore, if you want to draw a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this
radius through this point and this will be the required tangent at the point.

We are given a circle with center O and a point P outside it. We have to construct
the two tangents from P to the circle.


Steps of Construction:
1. Join PO and bisect it. Let M be the midpoint of PO.
2. Taking M as center and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R.
3. Join PQ and PR. Then PQ and PR are the required two tangents , now let us see how this construction works. Join OQ. Then ∠ PQO is an angle in the semicircle and, therefore,∠ PQO = 90° Can we say that PQ ⊥ OQ? Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.