PARABOLA 2 I NTRODUCTION In the previous lesson the parabola as a - PowerPoint PPT Presentation

D AY 137 – E QUATION OF A PARABOLA 2

I NTRODUCTION In the previous lesson the parabola as a curve was introduced, its prominent features like the focus and directrix were defined. Moreover, finding its equation when given the directrix and the focus was discussed. In several cases, the equation of the parabola is given, and we are required to find the equation of the directrix and the coordinates of the focus. The standard equations of the different types of parabolas are key when it comes to finding both the foci and directrixes. In this lesson will learn how to find the equation of the directrix and the coordinates of the focus when given the equation of a parabola.

V OCABULARY  A parabola The set of all points in a plane that are equidistant from a fixed point in the curve referred to as the focus and a fixed line outside the curve referred to as the directrix .

 Focal width of a parabola A line segment parallels to the directrix and passing through the focus of a parabola whose both endpoints are on the parabola. It is also referred to as the latus rectum of the parabola. It is given by the expression: 4𝑞 , where 𝑞 is the distance from the focus to the vertex.

T HE EQUATION OF A PARABOLA  The standard equation of a parabola whose focus, F is at 0, 𝑞 and the equation of its directrix is 𝑧 = −𝑞 is 𝒚 𝟑 = 𝟓𝒒𝒛 . If 𝑞 > 0 the parabola opens upward whereas if 𝑞 < 0 the parabola opens downward. The axis of symmetry is vertical, that is, the 𝑧 − 𝑏𝑦𝑗𝑡 .  The standard equation of a parabola whose focus, F is at 𝑞, 0 and the equation of its directrix is 𝑦 = −𝑞 is 𝒛 𝟑 = 𝟓𝒒𝒚 . If 𝑞 > 0 the parabola opens towards the right whereas if 𝑞 < 0 the parabola opens towards the left. The axis of symmetry is horizontal, that is, the 𝑦 − 𝑏𝑦𝑗𝑡 .

E QUATIONS OF TRANSLATED PARABOLAS  In a number of cases the equations represent parabolas that have been translated either parallel to the 𝑦 − 𝑏𝑦𝑗𝑡 or parallel to the 𝑧 − 𝑏𝑦𝑗𝑡 . The following equations represent general equations of parabolas.  The standard form of the equation of a parabola is 𝑦 − ℎ 2 = 4𝑞 𝑧 − 𝑙 , where the focus has coordinates ℎ, 𝑙 + 𝑞 and the directrix is given by the equation 𝑧 = 𝑙 − 𝑞 . The vertex is the point with coordinates ℎ, 𝑙 and the axis of symmetry is vertical.

 The standard form of the equation of a parabola is 𝑧 − 𝑙 2 = 4𝑞 𝑦 − ℎ , where the focus has coordinates ℎ + 𝑞, 𝑙 and the directrix is given by the equation 𝑦 = ℎ − 𝑞 . The vertex is the point with coordinates ℎ, 𝑙 and the axis of symmetry is horizontal.

T HE FOCUS AND THE DIRECTRIX FROM THE EQUATION OF A PARABOLA  The orientation of the parabola, that is, the direction it opens towards, helps us to identify the directrix correctly.  Finding the focus and the directrix of a parabola enables us to graph the parabola.  We use the geometric properties of the vertex, focus, directrix and the axis of symmetry together with the orientation to identify these parts from the equation.  Let us look at a few examples that demonstrate how to find the focus and the directrix from the equation of a parabola.

Example 1 Find the coordinates of the focus and the equation of the directrix of the parabola whose equation is 𝑦 2 = 24𝑧 Solution The equation 𝑦 2 = 24𝑧 is in the form 𝑦 2 = 4𝑞𝑧 . The axis of symmetry is vertical because the variable 𝑦 is squared. This shows that the parabola either opens upwards or downwards.

We then find the value of 𝑞 . 4𝑞 = 24 𝑞 = 6 𝑞 > 0 therefore the parabola opens upwards. Since the equation is in the form 𝑦 2 = 4𝑞𝑧 the coordinates of the focus will be in the form 0, 𝑞 ∴ The coordinates of the focus are 𝟏, 𝟕 . The equation of the directrix is of the form 𝑧 = −𝑞 . ∴ The equation of the directrix is thus 𝒛 = −𝟕

Example 2 The equation of a parabola is 𝑧 2 = 32𝑦 . Use this equation to find the equation of the directrix and the coordinates of the focus. Solution. The equation 𝑧 2 = 32𝑦 is in the form 𝑧 2 = 4𝑞𝑦 . The axis of symmetry is horizontal because the variable 𝑧 is squared. This shows that the parabola either opens towards the right or towards the left.

We then find the value of 𝑞 . 4𝑞 = 32 𝑞 = 8 𝑞 > 0 therefore the parabola opens towards the right. Since the equation is in the form 𝑧 2 = 4𝑞𝑦 the coordinates of the focus will be in the form 𝑞, 0 The coordinates of the focus are 𝟗, 𝟏 The equation of the directrix is of the form 𝑦 = −𝑞 . ∴ The equation of the directrix is thus 𝒚 = −𝟗

Example 3 Identify the focus and the directrix from the parabola 𝑧 − 2 2 = −4 𝑦 + 1 and hence find the focal width of the parabola. Solution The graph of the parabola is in the form: 𝑧 − 𝑙 2 = 4𝑞 𝑦 − ℎ The axis of symmetry of this parabola is horizontal. Using the equation, ℎ = −1 and 𝑙 = 2 . ∴ The vertex of the parabola is −1, 2

We then find the value of 𝑞 . Comparing the equations, we have: 4𝑞 = −4 𝑞 = −1 This shows that the parabola opens to the left. The focus is located 1 unit to the left of the vertex and has the coordinates ℎ + 𝑞, 𝑙 ∴ The coordinates of the focus are: −1 + −1 , 2 = −𝟑, 𝟑 The directrix is located 1 units to the right of the vertex and it is given by the equation 𝑦 = ℎ − 𝑞 ∴ The equation of the directrix is: 𝑦 = −1 − −1 ⇒ 𝒚 = 𝟏

The focal width of the parabola, also referred to as the latus rectum of a parabola is given by the expression: |4𝑞| In this case the focal width is given by: 4 × −1 = −4 = 4 ∴ The focal width is 4 .

 In other cases the equation of the parabola is not given in standard form. In such cases, we need to rewrite the equation of the parabola in standard form before embarking on the process of finding the focus and the directrix.

Example 4 Find the focus and the directrix of the parabola 𝑧 2 + 4𝑧 − 12𝑦 − 20 = 0 Solution This equation is not in standard form, we first convert it into the standard equation of a parabola of the form: 𝑧 − 𝑙 2 = 4𝑞 𝑦 − ℎ The equation has the variable 𝑧 squared, so we leave the terms with 𝑧 on the left hand side of the equation and take the other terms to the right hand side of the equation.

The equation becomes: 𝑧 2 + 4𝑧 = 12𝑦 + 20 We then complete the square on the left hand side of the equation. The equation thus becomes: 𝑧 2 + 4𝑧 + 4 = 12𝑦 + 20 + 4 𝑧 2 + 4𝑧 + 4 = 12𝑦 + 24 We then factorize the left hand side and simplify the right hand side. We therefore have: 𝑧 + 2 2 = 12 𝑦 + 2

We have written the equation in the standard form: 𝑧 − 𝑙 2 = 4𝑞 𝑦 − ℎ We can now find the focus and the directrix from the equation 𝑧 + 2 2 = 12 𝑦 + 2 We find the value of 𝑞 : 4𝑞 = 12 𝑞 = 3 The vertex is has coordinates ℎ, 𝑙 ∴ The vertex is the point −2, −2 The focus is 3 units to the right of the vertex with coordinates ℎ + 𝑞, 𝑙 .

∴ The coordinates of the focus are: −2 + 3, −2 = 𝟐, −𝟑 The directrix is 3 units to the left of the vertex with equation 𝑦 = ℎ − 𝑞 ∴ The equation of the directrix is: 𝑦 = −2 − 3 ⇒ 𝒚 = −𝟔

HOMEWORK Find the equation of the directrix and the coordinates of the focus of a parabola whose equation is 𝑧 − 4 2 = −16 𝑦 + 2 .

A NSWERS TO HOMEWORK The coordinates of the focus: −6, 4 The equation of the directrix: 𝑦 = 2

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