This is a powerpoint presentation that discusses about the topic or lesson: Conic Sections. It also includes the definition, types and some terminologies involved in the topic: Conic Sections.
This is a powerpoint presentation that discusses about the topic or lesson: Conic Sections. It also includes the definition, types and some terminologies involved in the topic: Conic Sections.
Atelier animé lors des journées de formation de l'Association Française des Fundraisers en juillet 2016, auprès d'une audience de représentants d'associations de divers secteurs.
Mon retour d'expérience de Chief Digital Officer sur la bonne démarche à adopter pour mener une transformation digitale efficace et qui fonctionne.
MS Report, When we talked about the conic section it involves a double-napped cone and a plane. If a plane intersects a double right circular cone, we get two-dimensional curves of different types. These curves are what we called the conic section.
9.2 - parabolas 1.ppt discussion about parabolassuser0af920
In mathematics, a parabola is a plane curve which is mirror-symmetrical and is approximately U-shaped. It fits several superficially different mathematical descriptions, which can all be proved to define exactly the same curves. One description of a parabola involves a point and a line. The focus does not lie on the directrix. The parabola is the locus of points in that plane that are equidistant from the directrix and the focus. Another description of a parabola is as a conic section, created from the intersection of a right circular conical surface and a plane parallel to another plane that is tangential to the conical surface. The graph of a quadratic function y=ax²+bx+c is a parabola if a≠0, and, conversely, a parabola is the graph of a quadratic function if its axis is parallel to the y-axis. The line perpendicular to the In mathematics, a parabola is a plane curve which is mirror-symmetrical and is approximately U-shaped. It fits several superficially different mathematical descriptions, which can all be proved to define exactly the same curves. One description of a parabola involves a point and a line. The focus does not lie on the directrix. The parabola is the locus of points in that plane that are equidistant from the directrix and the focus. Another description of a parabola is as a conic section, created from the intersection of a right circular conical surface and a plane parallel to another plane that is tangential to the conical surface. The graph of a quadratic function y=ax²+bx+c is a parabola if a≠0, and, conversely, a parabola is the graph of a quadratic function if its axis is parallel to the y-axis. The line perpendicular to the In mathematics, a parabola is a plane curve which is mirror-symmetrical and is approximately U-shaped. It fits several superficially different mathematical descriptions, which can all be proved to define exactly the same curves. One description of a parabola involves a point and a line. The focus does not lie on the directrix. The parabola is the locus of points in that plane that are equidistant from the directrix and the focus. Another description of a parabola is as a conic section, created from the intersection of a right circular conical surface and a plane parallel to another plane that is tangential to the conical surface. The graph of a quadratic function y=ax²+bx+c is a parabola if a≠0, and, conversely, a parabola is the graph of a quadratic function if its axis is parallel to the y-axis. The line perpendicular to the In mathematics, a parabola is a plane curve which is mirror-symmetrical and is approximately U-shaped. It fits several superficially different mathematical descriptions, which can all be proved to define exactly the same curves. One description of a parabola involves a point and a line. The focus does not lie on the directrix. The parabola is the locus of points in that plane that are equidistant from the directrix and the focus. Another description of a parabola is as a conic s
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)
Sol. First we find the equation of axis of parabola
Securing your Kubernetes cluster_ a step-by-step guide to success !KatiaHIMEUR1
Today, after several years of existence, an extremely active community and an ultra-dynamic ecosystem, Kubernetes has established itself as the de facto standard in container orchestration. Thanks to a wide range of managed services, it has never been so easy to set up a ready-to-use Kubernetes cluster.
However, this ease of use means that the subject of security in Kubernetes is often left for later, or even neglected. This exposes companies to significant risks.
In this talk, I'll show you step-by-step how to secure your Kubernetes cluster for greater peace of mind and reliability.
DevOps and Testing slides at DASA ConnectKari Kakkonen
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Transcript: Selling digital books in 2024: Insights from industry leaders - T...BookNet Canada
The publishing industry has been selling digital audiobooks and ebooks for over a decade and has found its groove. What’s changed? What has stayed the same? Where do we go from here? Join a group of leading sales peers from across the industry for a conversation about the lessons learned since the popularization of digital books, best practices, digital book supply chain management, and more.
Link to video recording: https://bnctechforum.ca/sessions/selling-digital-books-in-2024-insights-from-industry-leaders/
Presented by BookNet Canada on May 28, 2024, with support from the Department of Canadian Heritage.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
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This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
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Welcome to UiPath Test Automation using UiPath Test Suite series part 3. In this session, we will cover desktop automation along with UI automation.
Topics covered:
UI automation Introduction,
UI automation Sample
Desktop automation flow
Pradeep Chinnala, Senior Consultant Automation Developer @WonderBotz and UiPath MVP
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
2. Conic Sections - Parabola The intersection of a plane with one nappe of the cone is a parabola.
3. Conic Sections - Parabola The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.
4. Conic Sections - Parabola The line is called the directrix and the point is called the focus . Focus Directrix
5. Conic Sections - Parabola The line perpendicular to the directrix passing through the focus is the axis of symmetry . The vertex is the point of intersection of the axis of symmetry with the parabola. Focus Directrix Axis of Symmetry Vertex
6. Conic Sections - Parabola The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d 1 = d 2 for any point (x, y) on the parabola. Focus Directrix d 1 d 2
8. Conic Sections - Parabola We know that a parabola has a basic equation y = ax 2 . The vertex is at (0, 0). The distance from the vertex to the focus and directrix is the same. Let’s call it p . Focus Directrix p p y = ax 2
9. Conic Sections - Parabola Find the point for the focus and the equation of the directrix if the vertex is at (0, 0). Focus ( ?, ?) Directrix ??? p p ( 0, 0) y = ax 2
10. Conic Sections - Parabola The focus is p units up from (0, 0), so the focus is at the point (0, p). Focus ( 0, p) Directrix ??? p p ( 0, 0) y = ax 2
11. Conic Sections - Parabola The directrix is a horizontal line p units below the origin. Find the equation of the directrix. Focus ( 0, p) Directrix ??? p p ( 0, 0) y = ax 2
12. Conic Sections - Parabola The directrix is a horizontal line p units below the origin or a horizontal line through the point (0, -p). The equation is y = -p. Focus ( 0, p) Directrix y = -p p p ( 0, 0) y = ax 2
13. Conic Sections - Parabola The definition of the parabola indicates the distance d 1 from any point (x, y) on the curve to the focus and the distance d 2 from the point to the directrix must be equal. Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, y) y = ax 2 d 1 d 2
14. Conic Sections - Parabola However, the parabola is y = ax 2 . We can substitute for y in the point (x, y). The point on the curve is (x, ax 2 ). Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2
15. Conic Sections - Parabola What is the coordinates of the point on the directrix immediately below the point (x, ax 2 )? Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( ?, ?)
16. Conic Sections - Parabola The x value is the same as the point (x, ax 2 ) and the y value is on the line y = -p, so the point must be (x, -p). Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( x, -p)
17. Conic Sections - Parabola d 1 is the distance from (0, p) to (x, ax 2 ). d 2 is the distance from (x, ax 2 ) to (x, -p) and d 1 = d 2 . Use the distance formula to solve for p. Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( x, -p)
18. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax 2 ). d2 is the distance from (x, ax 2 ) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d 1 = d 2 You finish the rest.
19. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax 2 ). d2 is the distance from (x, ax 2 ) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d 1 = d 2
20. Conic Sections - Parabola Therefore, the distance p from the vertex to the focus and the vertex to the directrix is given by the formula
21. Conic Sections - Parabola Using transformations, we can shift the parabola y=ax 2 horizontally and vertically. If the parabola is shifted h units right and k units up, the equation would be The vertex is shifted from (0, 0) to (h, k). Recall that when “a” is positive, the graph opens up. When “a” is negative, the graph reflects about the x-axis and opens down.
22. Example 1 Graph a parabola. Find the vertex, focus and directrix.
23. Parabola – Example 1 Make a table of values. Graph the function. Find the vertex, focus, and directrix.
24. Parabola – Example 1 The vertex is (-2, -3). Since the parabola opens up and the axis of symmetry passes through the vertex, the axis of symmetry is x = -2.
25. Parabola – Example 1 Make a table of values. x y -2 -1 0 1 2 3 4 -3 -1 Plot the points on the graph! Use the line of symmetry to plot the other side of the graph.
30. Conic Sections - Parabola The latus rectum is the line segment passing through the focus, perpendicular to the axis of symmetry with endpoints on the parabola. y = ax 2 Focus Vertex (0, 0) Latus Rectum
31. Conic Sections - Parabola In the previous set, we learned that the distance from the vertex to the focus is 1/(4a). Therefore, the focus is at y = ax 2 Focus Vertex (0, 0) Latus Rectum
32. Conic Sections - Parabola Using the axis of symmetry and the y-value of the focus, the endpoints of the latus rectum must be y = ax 2 Vertex (0, 0) Latus Rectum
33. Conic Sections - Parabola Since the equation of the parabola is y = ax 2 , substitute for y and solve for x.
34. Conic Sections - Parabola Replacing x, the endpoints of the latus rectum are y = ax 2 Vertex (0, 0) Latus Rectum and
35. Conic Sections - Parabola The length of the latus rectum is y = ax 2 Vertex (0, 0) Latus Rectum
36.
37. Example 2 Graph a parabola using the vertex, focus, axis of symmetry and latus rectum.
38. Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph.
39. Parabola – Example 2 The vertex is at (1, 2) with the parabola opening down. The focus is 4 units down and the directrix is 4 units up. The length of the latus rectum is
40. Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph. V(1, 2) Directrix y=6 Focus (1, -2) Latus Rectum Axis x=1
41. Parabola – Example 2 The graph of the parabola V(1, 2) Directrix y=6 Focus (1, -2) Latus Rectum Axis x=1
42. x = ay 2 Parabola Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
43.
44.
45.
46.
47. x = a(y – k) 2 + h Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
48. Parabola – x = a(y – k) 2 + h When horizontal and vertical transformations are applied, a vertical shift of k units and a horizontal shift of h units will result in the equation: x = a(y – k) 2 + h We have just seen that a parabola x = ay 2 opens to the right when a is positive. When a is negative, the graph will reflect about the y-axis and open to the left. Note: In both cases of the parabola, the x always goes with h and the y always goes with k.
49. Example 3 Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
50. Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum. What is the vertex? Remember that inside the “function” we always do the opposite. So the graph moves -1 in the y direction and -2 in the x direction. The vertex is (-2, -1) What is the direction of opening? The parabola opens to the left since it is x= and “a” is negative.
51. Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum. What is the distance to the focus and directrix? The distance is The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola.
52. Parabola – Example 3 The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola. Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5
53.
54. Parabola – Example 3 Construct the latus rectum with a length of 2. Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5 Latus Rectum? 2 Construct the parabola.
55. Parabola – Example 3 The parabola is: Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5 Latus Rectum? 2
57. Table of Rules - y = a(x - h) 2 + k a > 0 a < 0 Opens Vertex Focus Axis Directrix Latus Rectum Up Down ( h , k ) ( h , k ) x = h x = h (h, k) (h, k) x = h x = h
58. Table of Rules - x = a(y - k) 2 + h a > 0 a < 0 Opens Vertex Focus Axis Directrix Latus Rectum Right Left ( h , k ) ( h , k ) y = k y = k (h, k) (h, k) y = k y = k
64. Example 4 – Satellite Receiver 8 ft 1 ft With a vertex of (0, 0), the point on the diameter would be (4, 1). Fit a parabolic equation passing through these two points. V(0, 0) (4, 1) y = a(x – h) 2 + k Since the vertex is (0, 0), h and k are 0. y = ax 2
65. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) y = a x 2 The parabola must pass through the point (4, 1). 1 = a (4) 2 Solve for a . 1 = 16 a
66. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) The model for the parabola is: The receiver should be placed at the focus. Locate the focus of the parabola. Distance to the focus is:
67. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) The receiver should be placed 4 ft. above the vertex.