problems on parabola
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The document contains 5 problems that involve finding the vertex, focus, axis, and directrix of different parabolas defined by equations of the form x^2 + Ax + By + C = 0. For each problem, the solution provides the vertex (focal point), focus, axis of symmetry, and directrix of the parabola defined by that equation.
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A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line.
1.1 Focus : The fixed point is called the focus of the Parabola.
1.2 Directrix : The fixed line is called the directrix of the Parabola.
(focus)
2.3 Vertex : The point of intersection of a parabola and its axis is called the vertex of the Parabola.
NOTE: The vertex is the middle point of the focus and the point of intersection of axis and directrix
2.4 Focal Length (Focal distance) : The distance of any point P (x, y) on the parabola from the focus is called the focal length. i.e.
The focal distance of P = the perpendicular distance of the point P from the directrix.
2.5 Double ordinate : The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola.
2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes through the focus of parabola then it is called as latus rectum.
2.7.1 Length of latus rectum :
The length of the latus rectum = 2 x perpendicular distance of focus from the directrix.
2.1 Eccentricity : If P be a point on the parabola and PM and PS are the distances from the directrix and focus S respectively then the ratio PS/PM is called the eccentricity of the Parabola which is denoted by e.
Note: By the definition for the parabola e = 1.
If e > 1 Hyperbola, e = 0 circle, e < 1
ellipse
2.2 Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola.
If we take vertex as the origin, axis as x- axis and distance between vertex and focus as 'a' then equation of the parabola in the simplest form will be-
y2 = 4ax
3.1 Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 𝑙 then coordinates of end points of this Double ordinate are
𝑙2 𝑙2
, 𝑙
and
, 𝑙 .
4a
4a
3.2 Other standard Parabola :
Equation of Parabola Vertex Axis Focus Directrix Equation of Latus rectum Length of Latus rectum
y2 = 4ax (0, 0) y = 0 (a, 0) x = –a x = a 4a
y2 = – 4ax (0, 0) y = 0 (–a, 0) x = a x = –a 4a
x2 = 4ay (0, 0) x = 0 (0, a) y = a y = a 4a
x2 = – 4ay (0, 0) x = 0 (0, –a) y = a y = –a 4a
Standard form of an equation of Parabola
Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13)
(B) (–13/4, –15/4)
(C) (15/2, – 13/2)
(D) (15/4, – 13/4)
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Parabola complete
This summary combines slides from Melanie Tomlinson and Morrobea on the topic of parabolas. The key points covered include:
- The geometric definition of a parabola as the set of all points equidistant from a fixed point (the focus) and fixed line (the directrix).
- Parabolas can be represented using various equation forms including vertex form, standard form, and general form.
- Methods for graphing parabolas by identifying features like the vertex, axis of symmetry, x-intercepts, focus, and directrix.
- Applications of parabolas to model real-world situations like searchlights and radio telescopes.
Graphing quadratics
1) To graph a quadratic function in standard form y=ax^2 + bx + c, one finds the axis of symmetry using the formula x=-b/2a, finds the vertex by substituting the axis into the function, and finds two other points to reflect across the axis and connect with a smooth curve.
2) The axis of symmetry always passes through the vertex. The y-intercept can be found by substituting x=0 into the function.
3) A quadratic function has either 0, 1, or 2 solutions which are found by setting the function equal to 0 and solving the resulting quadratic equation.
Review for the Third Midterm of Math 150 B 11242014Probl.docx
Review for the Third Midterm of Math 150 B 11/24/2014
Problem 1
Recall that 1
1−x =
∑∞
n=0 x
n for |x| < 1.
Find a power series representation for the following functions and state the radius of
convergence for the power series
a) f(x) = x
2
(1+x)2
.
b) f(x) = 2
1+4x2
.
c) f(x) = x
4
2−x.
d) f(x) = x
1+x2
.
e) f(x) = 1
6+x
.
f) f(x) = x
2
27−x3 .
Problem 2
Find a Taylor series with a = 0 for the given function and state the radius of conver-
gence. You may use either the direct method (definition of a Taylor series) or known
series.
a) f(x) = ln(1 + x)
b) f(x) = sin x
x
c) f(x) = x sin(3x).
Problem 3
Find the radius of convergence and interval of convergence for the series
∑∞
n=1
(x+2)n
n4n
.
Ans. Radius r=2,
√
2 − 2 < x <
√
2 + 2. Problem 4
Find the interval of convergence of the following power series. You must justify your
answers.
∑∞
n=0
n2(x+4)n
23n
.
Ans. −12 < x < 4.
Problem 5
For the function f(x) = 1/
√
x, find the fourth order Taylor polynomial with a=1.
Problem 6
A curve has the parametric equations
x = cos t, y = 1 + sin t, 0 ≤ t ≤ 2π
a) Find dy
dx
when t = π
4
.
b) Find the equation of the line tangent to the curve at t = π/4. Write it in y = mx+b
form.
c) Eliminate the parameter t to find a cartesian (x, y) equation of the curve.
d) Using (c), or otherwise identify the curve.
Problem 7
State whether the given series converges or diverges
a)
∑∞
n=0 (−1)
n+1 n22n
n!
.
b)
∑∞
n=0
n(−3)n
4n−1
.
c)
∑∞
n=1
sin n
2n2+n
.
Problem 8
1
Approximate the value of the integral
∫ 1
0
e−x
2
dx with an error no greater than 5×10−4.
Ans.
∫ 1
0
e−x
2
dx = 1 − 1
3
+ 1
5.2!
− 1
7.3!
+ ... +
(−1)n
(2n+1)n!
+ .... n ≥ 5,
for n=5
∫ 1
0
e−x
2
dx ≈ 1 − 1
3
+ 1
5.2!
− 1
7.3!
+ 1
9.4!
− 1
11.5!
≈ 0.747.
Problem 9
Find the radius of convergence for the series
∑∞
n=1
nn(x−2)2n
n!
.
Ans. R = 1√
e
.
Problem 10
Let f(x) =
∑∞
n=0
(x−1)n
n2+1
.
a) Calculate the domain of f.
b) Calculate f ′(x).
c) Calculate the domain of f ′.
Problem 11
Let f(x) =
∑∞
n=0
cos n
n!
xn.
a) Calculate the domain of f.
b) Calculate f ′(x).
c) Calculate
∫
f(x)dx.
Problem 12
Using properties of series, known Maclaurin expansions of familiar functions and their
arithmetic, calculate Maclaurin series for the following.
a) ex
2
b) sin 2x
c)
∫
x5 sin xdx
d) cos x−1
x2
e)
d((x+1) tan−1(x))
dx
Problem 13
Calculate the Taylor polynomial T5(x), expanded at a=0, for
f(x) =
∫ x
0
ln |sect + tan t|dt.
Ans. T5(x) =
x2
2
+ x
4
4!
.
Problem 14
Suppose we only consider |x| ≤ 0.8. Find the best upper bound or maximum value
you can for∣∣∣sin x − (x − x33! + x55! )∣∣∣
Same question: If
(
x − x
3
3!
+ x
5
5!
)
is used to approximate sin x for |x| ≤ 0.8. What is
the maximum error? Explain what method you are using.
Problem 15
The Taylor polynomial T5(x) of degree 5 for (4 + x)
3/2 is
(4 + x)3/2 ≈ 8 + 3x + 3
16
x2 − 1
128
x3 + 3
4096
x4 − 3
32768
x5.
a) Use this polynomial to find Taylor polynomials for (4 + ...
Grph quad fncts
The document discusses graphing quadratic functions. It defines a quadratic function as f(x) = ax^2 + bx + c where a, b, and c are real numbers and a is not equal to 0. The graph of a quadratic function is a parabola that is symmetrical about an axis. When the leading coefficient a is positive, the parabola opens upward and the vertex is a minimum. When a is negative, the parabola opens downward and the vertex is a maximum. Standard forms for quadratic functions and methods for finding characteristics like the vertex, axis of symmetry, and x-intercepts from the equation are also presented.
Parabola direction , vertex ,roots, minimum and maximum
This document discusses quadratic functions of the form f(x) = ax^2 + bx + c. It provides details on:
1) The direction a parabola opens depending on whether a is positive or negative.
2) How to calculate the vertex (turning point) of a parabola using the formula (-b/2a, 4ac-b^2/4a).
3) How to calculate the roots (x-intercepts) of a quadratic function using the formula (-b ± √(b^2 - 4ac))/2a.
Several examples are provided to demonstrate finding the direction, vertex, roots, minimum/maximum values of various quadratic functions.
Analytic geometry lecture2
This document discusses key concepts about circles, including:
- The standard equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) are the coordinates of the center and r is the radius.
- Given the equation or properties of a circle, one can determine its center and radius or write the equation in standard form.
- Points can lie inside, outside, or on a circle, which can be determined by comparing distances or substituting into the equation.
- A circle and line can intersect in 0, 1, or 2 points, which can be found using algebraic techniques.
- The equation of a circle can be found given 3
Straight-Line-Graphs-Final -2.pptx
This document contains 5 math problems involving factorizing expressions, solving equations, evaluating expressions for given values, expanding expressions, and finding the highest common factor. It also provides context on working with straight line graphs, including finding the gradient and y-intercept of a line from its equation, finding the gradient between two points, finding the midpoint and a point that divides a line segment in a given ratio, and finding the x- and y-intercepts of a line.
Notes parabolas
The document discusses the properties of parabolas that open up or down and parabolas that open left or right. It provides the general form and standard form of each, noting that a positive value of a in the standard form indicates the direction the parabola opens (up if discussing vertical parabolas, right if discussing horizontal parabolas) while a negative a indicates the opposite direction. It also identifies the axis of symmetry formula for each. Examples are given to demonstrate finding the direction a parabola opens, its vertex, and axis of symmetry.
Solving Quadratic Equations
This document discusses solving quadratic equations. It explains the difference between expressions and equations, and between linear and quadratic expressions. It then outlines the steps to solve a quadratic equation: set it equal to 0 if it's not already, factor the resulting expression, set each factor equal to 0 and solve for the variable, and check the solutions in the original equation. Several examples are worked through as practice problems.
LApreC2010-106matrixaprtakehome_4r
This document contains an exam on polynomial zeros and polygonal areas with several theorems and questions. It begins by covering theorems for solving quadratic, cubic, and quartic equations with complex coefficients using formulas like the quadratic formula. It then presents questions to demonstrate solving these types of equations. The second part introduces a theorem for calculating the area of any polygon by breaking it into triangles and using vector cross products, along with questions to show this relates to triangulation.
10.4
Here are the steps to solve each equation by graphing:
1) x2 + 5x + 6 = 0
Write in standard form: x2 + 5x + 6 = 0
Graph the related function y = x2 + 5x + 6. The x-intercepts are the solutions, which are -3 and -2.
2) x2 + 8x + 16 = 0
Write in standard form: x2 + 8x + 16 = 0
Graph the related function y = x2 + 8x + 16. The x-intercepts are the solutions, which are -4 and -4.
3) x2 - 2x + 3 = 0
Write in standard
Similar to problems on parabola (20)
Review for the Third Midterm of Math 150 B 11242014Probl.docx
Review for the Third Midterm of Math 150 B 11242014Probl.docx
Parabola direction , vertex ,roots, minimum and maximum
Parabola direction , vertex ,roots, minimum and maximum
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problems on parabola
- 2. Problem 1: Find the vertex, focus, axis, and directrix of the following parabola: x2 - 4x - 8y + 28 = 0. Solution for Problem 1 Vertex: (2, 3). Focus: (2, 5). Axis: x = 2. Directrix: y = 1. Problem 2: Find the vertex, focus, axis, and directrix of the following parabola: y2 + 4x - 2y - 3 = 0. Solution for Problem 2 Vertex: (1, 1). Focus: (0, 1). Axis: y = 1. Directrix: x = 2. Problem 3: Find the vertex, focus, axis, and directrix of the following parabola: x2 - 4y = 0. Solution for Problem 3 Vertex: (0, 0). Focus: (0, 1). Axis: x = 0. Directrix: y = - 1.
- 3. Problem 4: Find the vertex, focus, axis, and directrix of the following parabola: x2 - 10x - 2y + 29 = 0 Solution for Problem 4 Vertex: (5, 2). Focus: (5, 2.5). Axis: x = 5. Directrix: y = 1.5. Problem 5: Find the vertex, focus, axis, and directrix of the following parabola: y2 - 16x + 16 = 0. Solution for Problem 5 Vertex: (0, 1). Focus: (4, 1). Axis: y = 1. Directrix: x = - 4.