History BC Dating back to the 380s, the parabola has always been one of the most striking areas of mathematics. Unlike other subjects, the parabola was not invented. On the contrary, it was discovered and various studies were conducted on it. Nowadays, it is used in many practical applications.
In this article, we explained the subject by discussing the parabola in all its details. Additionally, in order to reinforce the subject, we solved sample questions that appeared in previous exams using frequently used equations. Without further ado, let’s move on to our content.
Let’s start from the basics; What is a parabola?
A parabola is simply the equation of a particular curve; so that every point on the curve is always equidistant from a fixed point and a fixed line. This fixed point is the focus of the parabola and the fixed line is the directrix of the parabola. In other words, the geometric part of a point that is equidistant from a given point or a given line is called a parabola.
Parabolic curves are widely used in many fields such as physics, engineering, finance and computer science. A parabola is a U-shaped curve that can be concave up or down depending on the equation.
So how did the parabola come about?
Historically, the geometric properties of the parabola were revealed by the ancient Greeks. When historical records are examined, Menacchmus (380-320 BC) appears as the first person to examine the properties of parabola as well as hyperbola and ellipse. Later, Apollonius of Perga B.C. 262-190) made a study on conic sections, but his definition of parabola was in terms of ratios and he did not use any coordinates in his work.
Many years later, Galileo (1564-1642) noticed that a bullet moves in a parabolic path under the influence of gravity. The famous mathematician Kepler (1571-1630) was the first to notice that the planets rotate very close to an elliptical shape as they move around the Sun. Newton proved this using the law of universal gravitation.
What is the standard equation of a parabola and how is it written?
Take a point P with coordinates (x,y) on the parabola in the XY plane. By definition of a parabola, the distance of any point on the parabola from the focus and the directrix is equal. The distance of P to the directrix is expressed as PB; where the coordinates of B are (-a, y) since it is located on the direction and the distance of P from the focus is PF.
Since PF = PB (1) according to the definition of parabola, using the distance formula we obtain the following result:
PF = √((xa)² +(y-0)² = √{(xa)² +y² } . . . .(2)
PB = √{(x+a)² } (3)
(1), (2) and (3). Using the equations we get the following result:
√{(x−a) +y² } = √{(x+a)² }
⇒ (xa)² + y² = (x+a)²
⇒ x² + a² – 2ax + y² = x² + a² + 2ax
⇒ y² – 2ax = 2ax
= y² = 4ax
In general, in the standard equation of a parabola, if the direction is parallel to the y-axis, an equation is formed as follows;
y² = 4ax
If the parabola is side by side, that is, if the direction is parallel to the x-axis, the standard equation of the parabola becomes;
x² = 4months
Apart from these two cases, the equation of a parabola can be y² = -4ax and x² = -4ay if the parabola is in the negative part.
What are the parabola formulas?
Formula of a parabola with known vertex and one point
y = a.(xr)² + k
Parabola formula where the points intersected by the x-axis and another point on it are known
f(x) = a. (x – x1) . (x – x2)
Formula for a parabola with three known points
y=f(x) =ax² + bx + c
Sample parabola questions in exams
Question 1:
The vertex of the parabola whose graph is shown in the figure is T(5/2, 5) and the point where it intersects the y-axis is A(0,4). Since the equation of this parabola is y = ax² + bx + c, what is b?
Solution 1:
Using vertex,
We can write the equation as y = a(x+5/2)² + 5.
Let’s also use the point (0,4).
4 = a(0+5/2)² + 5
4 = 25a / 4 + 5
-1 = 25a/4 => -4 = 25a => a = – 4 / 25.
Let’s expand and write the equation y = – 4 / 25 (x+5/2)² + 5.
= – 4 / 25 x² + 5x + 25/4) + 5
= – 4 /25 x² – 4 / 5 x +.
Accordingly, b = -⅘.
Question 2:
What is the maximum value that the expression x² + y² can take for the (x, y) points on the borders of the limited region between the y = x² parabola and the y = 2 – x line?
Solution 2:
First, let’s draw the graph of the parabola and the line. The y = x² parabola is a parabola whose vertex is the origin and its arms are upward. The y = 2 – x line intersects the axes at points 2 and 2.
For the largest value of x²+ y², we should take the point furthest from the origin. This is point A. Let’s find the coordinates of point A.
x²- 2 – x
x² + x – 2 = 0
(x+2)(x-1) = 0 , x = -2 is the abscissa of A.
For x = -2, y = 2 – x – 4.
For point A(-2, 4)
x² + y² = (-2)² + 4²
= 4 + 16 = 20.
Question 3:
y = x² – 2(a + 1 )x + a² -1
Since the parabola is tangent to the line y = 1, what is a?
Solution 3:
When y = 1, the function must have a single root so that it is tangent to the line y = 1. In that case,
1 = x² – 2(a+1)x + a² -1
In the equation 0 = x² -2(a +1)x + a² – 2, Δ = 0.
(-2(a+1))² – 4.a.(a²-2)=0
4(a² + 2a + 1)- 4a² + 8 = 0
8a + 12 = 0
8a = -12
a= -3/2
Question 4:
Passing through the origin in the perpendicular coordinate plane, where a and b are positive real numbers,
p (x + a) + b
p (x + a) – b
p (x – a) – b
The vertices of the three parabola defined as are the vertices of a triangle with an area of 16 square units.
Accordingly, what is the sum of a + b?
Solution 4:
If the polynomial p (x) = (xa)² -b passes through the origin, the point (0,0) is a point of the parabola.
0 = (0-a)² – b
0 = a² – b => a² = b.
Let’s find the vertex of other polynomials.
p ( x + a) + b = (((x+a)-a)² – b ) + b
=x² – b + b
= x². The vertex is the (0,0) point.
p ( x + a) -b = (((xa)-a)² -b ) -b
= x²-bb
= x² – 2b. The peak is point (0,2b).
p (xa) -b =(((xa)-a)² – b ) -b
= (x-2a)² -b -b
= (x-2a)² -2b. The vertex is point (2a, -2b).
Let’s examine it on the coordinate plane.
A triangle is obtained whose base is 2a long and whose height is 2b long. If the area of this triangle is 16br²,
(2b.2a)/2 = 16
ab = 8.
If a² = b then a = 2.
b = a² = 4.
a + b = 2+4 = 6
Question 5:
The graph of the function f(x) is a parabola that is tangent to the Ox axis at the point (1, 0) and passes through the point (0, 3), as shown in the figure.
Accordingly, what is f(3)?
Solution 5:
If the vertex is (1, 0) point
The equation y = a(x-1)² + 0 emerges.
If we assume that it passes through the point (0, 3),
3 = a(0-1)²
3 = a.1 => a = 3
If f(x) = 3(x-1)² then f(3)=3.(3-1)² = 3.4 = 12.
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