Sketching a Parabola In Exercises 11-16, find the vertex, focus, and directrix of the parabola,

Sketching a Parabola In Exercises 11-16, find the vertex,...

Key Concepts

Focus and Directrix of a Parabola

The focus is a fixed point that lies inside the parabola, while the directrix is a line located outside the curve. The definition of a parabola involves all points on the curve being equidistant from the focus and the directrix, making these components fundamental to the geometry of the parabola.

Parabolic Equation in Standard Form

A parabolic equation can be rewritten in a standard form to clearly show its orientation and key parameters, such as the vertex, focus, and directrix. Recognizing whether the parabola opens vertically or horizontally is essential for correctly interpreting the equation.

Completing the Square

Completing the square is an algebraic method used to convert a quadratic expression into a perfect square form. This technique is particularly useful for rewriting a parabolic equation into vertex form, which makes it easier to identify the vertex and subsequently the focus and directrix.

Vertex of a Parabola

The vertex is the point on the parabola where it attains its maximum or minimum value, and it serves as the axis of symmetry for the curve. In vertex form, the coordinates of the vertex are directly visible, facilitating further analysis and graphing of the parabola.

Graphing a Parabola

Graphing a parabola involves plotting its vertex, drawing its axis of symmetry, and marking additional elements such as the focus and directrix. A well-drawn graph visually represents the properties of the quadratic function and aids in understanding the spatial relationships between these key elements.

Transcript

00:01 Hi guys.

00:03 In this question, we're going to be looking at the equation of a parabola.

00:09 So let's type everything.

00:11 We are given an equation of a parabola, parabola of the following form.

00:22 So the form of the equation is as follows, is y squared minus 4y minus 4x equals to 0.

00:33 Now let's give this equation and number, it's customary to do.

00:38 So when we say, of course, an equation of a parabola, right, we understand, of course, this as that we have a parabola in the r2 plane.

00:48 It is a set of points in the r2 plane that satisfy certain distant relation with respect to its focus point in the directrics and also satisfy this equation.

00:59 And the question asks us to find three things.

01:03 So we would like to find number one the vertex, number two we would like to find the focus, and number three we would like to find the directrix of this parabola.

01:21 How do we do this? we can try and make use of the theorem 10 .1 from your book, and i will state this theorem in a form that it will be convenient for us to use.

01:33 And then we're going to try and apply this theorem to find these three things.

01:37 And once we've done that, we can also try to sketch the graph of this parabola.

01:43 So let's, on this new page, let's state the theorem, the theorem that we're going to be using.

01:51 Let me write down the theorem like that, and the theorem is as follows.

01:58 If a parabola is given by an equation of the form, of the following form, let me switch to my pan so if it's given by an equation of the form y minus k squared equals p 4p times x minus h where where k p and h are some real numbers if this is true then the following conclusion is also true so we will say then, and what is true? let me switch back to the pen.

02:52 So then h plus p, k is the focus of the parabola, is the focus.

03:09 Hk is the vertex, and an x equals h minus p is the line, which is the directrix of the parabola.

03:29 Is the directrix.

03:35 So we're going to try to use this theorem in order to produce an answer to our question.

03:40 How can this theorem be used? well, if we look back at our initial problem, so we are told that we're given a problem that satisfies this equation, right? which, of course, immediately does not look like the equation it would need to satisfy in order for us to apply the theorem.

03:56 But all is not lost.

03:57 We can always try and see if we can arrange the equation into a suitable form into this so -called standard form.

04:03 And if we can, then the conclusion will be immediate from this theory.

04:10 So let's bring up a new page and see if we can do some work on our equation and arrange this into a suitable.

04:17 So let me start by rewriting our equation.

04:21 So our equation looks like that.

04:23 Y squared minus 4 .y minus 4x equals to 0.

04:27 I would like to complete the square on the ys over here.

04:35 So if i complete the square, i will arrive at equation of the following form, right? it's going to be minus 4, minus 4x equals to 0.

04:47 You can check by expanding the square that the equation hasn't changed.

04:52 And i will proceed by moving 4 and 4x on the other side of this equation.

05:00 And also taking four outside of the bracket, i can arrive at an equation of the following form.

05:09 And as a final touch or final simplification, i would like to write this also as like that in the following form...

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