Parabola – vertex, focus, directrix, latus rectum

by Namagwa Benard
(Kenya)

Sideways, or Horizontal, Parabola

• Given

4y² - 8y + 3x - 2 = 0

• Show

the equation represents a parabola

• Find the:

vertex

focus

directrix

length of latus rectum

Comments for Parabola – vertex, focus, directrix, latus rectum

 Oct 14, 2012 Sideways Parabola by: Staff Answer: Part I 4y² - 8y + 3x - 2 = 0 represents a sideways, or horizontal, parabola.          Solve for x `4y² - 8y + 3x - 2 = 0 ` ```Subtract 4y² from both sides of the equation 4y² - 8y + 3x - 2 - 4y² = 0 - 4y² 4y² - 4y² - 8y + 3x - 2 = 0 - 4y² 0 - 8y + 3x - 2 = 0 - 4y² - 8y + 3x - 2 = - 4y² Add 8y to both sides of the equation - 8y + 3x - 2 + 8y= - 4y² + 8y - 8y + 8y + 3x - 2 = - 4y² + 8y 0 + 3x - 2 = - 4y² + 8y 3x - 2 = - 4y² + 8y Add 2 to both sides of the equation 3x - 2 + 2 = - 4y² + 8y + 2 3x + 0 = - 4y² + 8y + 2 3x = - 4y² + 8y + 2 Divide each side of the equation by 3 3x / 3 = (- 4y² + 8y + 2) / 3 x * (3 / 3) = (- 4y² + 8y + 2) / 3 x * (1) = (- 4y² + 8y + 2) / 3 x = (- 4y² + 8y + 2) / 3 x = (-4/3)y² + (8/3)y + 2/3 ```          Plot the equation -------------------------------------------------

 Oct 14, 2012 Sideways Parabola by: Staff ------------------------------------------------- Part II The vertex of the parabola is (2,1)          You can solve for the vertex of the parabola using the first term of the quadratic equation. ```the quadratic formula is: x = (-b/2a) ± [√(b² - 4ac)/2a] Because this is a sideways parabola, the x and y variables must be reversed. y = (-b/2a) ± [√(b² - 4ac)/2a] the x coordinate of the vertex of the parabola is: y_vertex = (-b/2a) for your equation: x = (-4/3)y² + (8/3)y + 2/3 a = (-4/3) b = (8/3) y_vertex = -(8/3) / [2*(-4/3)] y_vertex = -(8/3) / (-8/3) y_vertex = 1 now that you know the y coordinate of the vertex, substitute this value in the original equation to compute x x = (-4/3)y² + (8/3)y + 2/3 x_vertex = (-4/3)*1² + (8/3)*1 + 2/3 x_vertex = (-4/3)*1 + (8/3)*1 + 2/3 x_vertex = (-4/3) + (8/3) + 2/3 x_vertex = (-4 + 8 + 2)/3 x_vertex = (6)/3 x_vertex = 2 the coordinates of the vertex in x,y format are: (2, 1) ```          You can also compute the vertex by rewriting the equation in vertex format as follows: ``` x = (-4/3)y² + (8/3)y + 2/3 x = [(-4/3)y² + (8/3)y] + 2/3 x = [(-4/3)y² + 2*(4/3)y] + 2/3 x = (-4/3)[y² - 2*1y ] + 2/3 x = (-4/3)[y² - 2*1y + 1 - 1 ] + 2/3 x = (-4/3)[y² - 2*1y + 1] + 4/3 + 2/3 x = (-4/3)(y - 1)² + 4/3 + 2/3 x = (-4/3)(y - 1)² + (4 + 2)/3 x = (-4/3)(y - 1)² + (6)/3 x = (-4/3)(y - 1)² + 2 As you can see, the coordinates of the vertex in x,y format are: (2, 1) ``` The focus of the parabola is (29/16, 1) -------------------------------------------------

 Oct 14, 2012 Sideways Parabola by: Staff ------------------------------------------------- Part III ```The Standard equation of a parabola with a horizontal axis is: (y - k)² = 4p(x - h) The coordinates of the vertex is (h,k) Substituting the values for the vertex (2, 1) for your equation: (y - 1)² = 4p(x - 2) To solve for p, enter in a point on the curve, such as (2/3, 0). when y = 0, x = 2/3 x = (-4/3)y² + (8/3)y + 2/3 x = (-4/3)*0² + (8/3)*0 + 2/3 x = 0 + 0 + 2/3 x = 2/3 (0 - 1)² = 4p(2/3 - 2) (- 1)² = 4p(2/3 - 6/3) 1 = 4p(- 4/3) 1 = p(- 16/3) 1 * (-3/16) = p(- 16/3) * (-3/16) -3/16 = p * 1 -3/16 = p p = -3/16 the final equation is: (y - 1)² = 4 * -3/16 * (x - 2) (y - 1)² = (-3/4) * (x - 2) Since p = -3/16, the focus is 3/16 units to the left of the vertex. The focus = (2 - 3/16, 1) = (32/16 - 3/16, 1) = (29/16, 1) the coordinates of the focus in x,y format are: (29/16, 1) ``` The directrix of the parabola is x = 35/16 ```Since p = -3/16, the directrix is 3/16 units to the right of the vertex. The directrix = 2 + 3/16 = 32/16 + 3/16 = 35/16 the x coordinate of the directrix is: 35/16 ``` The length of the latus rectum of the parabola is 3/4 ```length of latus rectum = 4|p| length of latus rectum = 4|-3/16| = 12/16 = 3/4 the length of latus rectum is 3/4 ``` Thanks for writing. Staff www.solving-math-problems.com

 Jan 14, 2021 x NEW by: Anonymous y^2+8x-6y+25=0

 Mar 25, 2021 What is this? 2003? NEW by: Anonymous I dont got time for this i need a real calculator a**holes.

 Sep 15, 2021 Parabola – vertex, focus, directrix, NEW by: Anonymous y^2-6y-8x+17=0