How to find the focus and directrix of a parabola using TI 89 calculator?
Haven't got a TI89 but the manual calculation isn't too difficult. x² + 6x + 2y + 11 = 0 x² + 6x = -2y -11 . . . .complete the square on the LHS (x + 3)² -9 = -2y - 11 (x + 3)² = -2y - 2 = -2(y + 1) = 4*-½(y + 1). The equation is now in the form (x - a)² = 4c(y - b) where the vertex is at (a,b) and the focus is at a distance of c from the vertex with the directrix being equidistant but diametrically opposed. Vertex (-3,-1) as c = -½ then Focus (-3½, -1) and the Directrix is y = -½.
Help finding the directrix, focus, and roots of a parabola?
Given the parabola y = x² - 5x + 5 Complete the square. y = x² - 5x + 5 y - 5 + 25/4 = (x² - 5x + 25/4) y + 5/4 = (x - 5/2)² The vertex (h,k) = (5/2, -5/4). The vertex form of a parabola with vertex (h,k) is 4p(y - k) = (x - h)² For the given parabola we have y + 5/4 = (x - 5/2)² 4p = 1 p = 1/4 The directed distance from the vertex to the focus is p. The focus is (h, k+p) = (5/2, -5/4 + 1/4) = (5/2, -1). The directed distance from the vertex to the directrix is -p. The directrix is y = k - p = -5/4 - 1/4 y = -3/2 For the zeros of the parabola set it equal to zero. y = x² - 5x + 5 0 = x² - 5x + 5 x = [5 ± √(25 - 4*1*5)] / 2 = (5 ± √5)/2
Finding the vertex, focus, and directrix of a parabola?
Put the equation of the parabola into the form 4p(x - h) = (y - k)² The vertex is (h,k) and the distance from the vertex to the focus is p. The distance from the vertex to the directrix is also p. For the question at hand we have: y² + 6y + 8x + 25 = 0 8x + 25 = -y² - 6y 8x + 25 - 9 = -(y² + 6y + 9) 8x + 16 = -(y + 3)² -8(x + 2) = (y + 3)² The vertex (h,k) is (-2,-3). Since the y term is squared the parabola opens sideways. Since the y squared term and the x term have opposite signs and are on the opposite side of the equal sign, the parabola opens to the left. The line of symmetry is a horizontal line thru the vertex. y = -3 4p = -8 p = -2 The focus is a distance | p | = 2 from the vertex along the line of symmetry. It is inside the parabola. The focus is (h + p, k) = (-2 - 2, -3) = (-4, -3) The directrix is also a distance | p | = 2 from the vertex but in the opposite direction. It is a line perpendicular to the line of symmetry and is outside the parabola. The directrix is a vertical line with the equation: x = h - p = -2 + 2 = 0 x = 0
How can I find the focus and directrix of the parabola, [math]y=x^2+4x+4[/math]?
The focus is the midpoint of the latus rectum.The latus rectum is four times as long as the distance from the focus to the vertex.The vertex is on the axis of symmetry.The axis of symmetry is perpendicular to the directrix.The vertex is as far away from the focus as the point where the directrix intersects the axis of symmetry.So…The vertex is (-2,0). It’s on the line y=(1/2)x+b. b is therefore 1. Why is the gradient 1/2? Check statement #2 of all statements made. (1/2)*4=2, for every 2 unit x progresses, y progresses by 1.Find the other point of intersection of y=(1/2)x+1 and y=x^2+4x+4.x^2+(7/2)x+3=0=(1/2)(x + 2)(2x + 3)The other x is -1.5. Plug in -1.5 and y is 0.25.The focus, being collinear with both (-1.5,0.25) and (-2,0), is (-2,0.25).The directrix is parallel to the latus rectum and a point on the directrix is 0.25 lower than the vertex, y-wise. The latus rectum has a gradient of 0, meaning the equation relies y being constant. The directrix is y=-0.25.Hint: imagine two squares (and maybe a circle) on a line to find the directrix, the focus, and the latus rectum.
Finding vertex, focus and directrix of parabola this given equation:?
y^2 + 2y + 1 = 8x + 7 + 1 (y+1)^2 = 8x + 8 8(x+1) = (y+1)^2 8 = 4c , distance from vertex to focus is c=2. Parabola opens right since y is squared and 8 is positive. Vertex (-1, -1) Focus: (1, -1) (2 places right of vertex) directrix: x=-3 (2 places left of vertex, vertical line)
Help finding the focus, directrix, vertex and focal width of a parabola?
2. (x+2)2=4(y-3) x^2+4x+4 = 4y -12 y = 1/4x^2+x+4 ax²+bx+c = 1/4x²+x+4 Standard form: y = a(x-h)²+k = 1/4(x-(-2))²+3 Axis of symmetry: x=-2; Vertex (minimum): (-2,+3) Focus: (-2,4); Directrix: y = 2 Quadratic formula: x = (-b ±√(b²-4ac)/(2a) = (-1±√(1-4)/(1/2) negative discriminant, two complex roots: -2±3.4641016151377544i
The focus of a parabola is (-3,5) and the directrix is y = ‒1. Which equation represents a parabola that is congruent to the given parabola?
Let’s start from the basics.Let [math]P(x, y)[/math] be a point on the locus. Suppose the focus is the point [math]S(-3,5)[/math]. Draw a perpendicular from [math]P(x, y)[/math] to the directrix, so as to meet the directrix at the point [math]M.[/math]The distance [math]PS[/math] can be written down with the help of the Distance Formula:[math]\displaystyle PS = \sqrt{(x+3)^2+(y-5)^2}[/math][math]\displaystyle PS^2 = (x+3)^2+(y-5)^2 \ldots (1)[/math]The perpendicular distance of the point [math]P[/math] from the directrix [math]y = -1[/math], is the length of segment [math]PM[/math]:[math]\displaystyle PM = |y+1|[/math][math]\displaystyle PM^2 = (y+1)^2 \ldots (2)[/math]We are given that the locus is a parabola. By definition of parabola, the distance [math]PS[/math] of [math]P(x, y)[/math] from the point [math]S(-3, 5)[/math] should be equal to the distance [math]PM[/math] of point [math]P[/math] from the directrix, i.e.[math]\displaystyle PS = PM[/math]Squaring both sides,[math]\displaystyle PS^2 = PM^2[/math]From (1) and (2),[math]\displaystyle (x+3)^2+(y-5)^2 = (y+1)^2[/math][math]\displaystyle \therefore y = \frac{1}{12}(x^2+6x+33)[/math]is the required equation of the parabola.
Find the vertex, focus, directrix, and focal width of the parabola. x = 10y2?
Vertex will be (0,0), and the parabola will open to the right. The standard form will be x/10 = y^2, so we know that 4p = 1/10, and that p = 1/40. Your directrix will be x=0-p, so the directrix is x=-1/40 The focus is (x+p, y), so the focus is (1/40, 0) At x = 1/40, y^2 = 1/400, so y = +/- 1/20. So 1/2 your width is 1/20. 2*(1/20) = 1/10 is your width at the focus.
How do I find the equation of a parabola if focus and directrix are given?
Above is the general equation of a vertical parabola whose focus and direction is given.You can plug in values of focus (a, b) and k to get the final equation of a particular parabola.Using similar logic and steps you can come up with the general equation of a horizontal parabola, where directrix is given by x = k and focus is again at (a, b).Hope this helps! :-)