Identify the focus and directrix of the parabola y=-15x^2, 6x^2-3y=0, 2y^2+10x=0!?
a vertical parabola 4p(y − k) = (x − h)^2 where: the vertex is at (h,k) the focus is at (h,k+p) the directrix is y = k−p a horizontal parabola 4p(x − h) = (y − k)^2 where: the vertex is at (h,k) the focus is at (h+p,k) the directrix is x = h−p y = -15x^2 ⇒ (-1/15)(y − 0) = (x − 0)^2 ⇒ 4p = -1/15 ⇒ p = -1/60 the focus is at (0,0+(-1/60)) = (0,-1/60) the directrix is y = 0−(-1/60) = y = 1/60 6x^2 − 3y = 0 ⇒ (1/2)(y − 0) = (x − 0)^2 ⇒ 4p = 1/2 ⇒ p = 1/8 the focus is at (0,0+1/8) = (0,1/8) the directrix is y = 0−(1/8) = y = -1/8 2y^2 + 10x = 0 ⇒ -5(x−0) = (y−0)^2 ⇒ 4p = -5 ⇒ p = -5/4 the focus is at (0+(-5/4),0) = (-5/4,0) the directrix is x = 0−(-5/4) = x = 5/4
Finding the vertex, focus, and directrix of a parabola?
The initial step is to define what these terms mean. For a general quadratic we have the equation which defines a parabola. You should draw this on a graph to help you. a*x^2 b*x c = 0 vertex aka the point of greatest curvature. This may be found by taking the first derivatie and setting it to 0. Meaning it is where the rate of change of the curve hits 0 because the change is already at its highest.. That will be 2*a*x b = 0 => x = -b/(2*a). focus aka a single point on the axis of symmetry. This and the directrix may be used to define a parabola aka generate the quadratic equation you got. The parabola i.e. the graph of your equation is such that at any point the sum of shortest distance of the point from focus directerix is the same. directricx is a line perpendicular to the axis of symmetry and may be represented by x = p. For your equation re write it as the parabola below 2*y = x^2 - 4*x y = x^2/2 - 2*x a = 1/2 b = -2 c = 0 The vertex is at x = b/2a and y as defined by the equation so for vertex x = -b/2a = 2/1 = 2; y = 1/2 * 2^2 - 2*2 = 2-4 = -2 The focus goes throught the axis of symmetry and the vertex so it must have x =-b/2a = 2; we do not yet know the y. The directrix is a line so we just need to figure out its y. We denote the y for the foucs as "Y". The distance from the parabola to focus on the y axis is Y - 2. The distance from parabola to focus on the x axis is -b/2a or 2/2 = 1 Plug in Y - 2 = 1 => Y = 3 hence the focus is at x=2; y = 3 Now for directrix we need to find its y on the axis of symmetry. There the distance of directrix from parabola is the same as that of the focus. y(parabola) - y(directrix) = y(focus) - y(parabola) => -2 - y(directrix) = 3 - 1 => y(directrix) = 2 - 2 directrix is x = 0
Find the focus, directrix, and focal diameter of the parabola. x2 = 2y?
x^2 = 2y Here the equation of the parabola is of the form x^2 = 4py where (0, p) is the focus. 2y = 4py 2 = 4p p = 1/2 So the focus is (0, 1/2). Thus, the directrix is y = -1/2. The focal diameter is 2(1/2) = 1.
How do I find the focus and directrix of the equation of the parabola y^2-6x+2y+13=0?
I have hardly ever needed to use “focus” and “directrix” of conics in over 45 years of teaching. It seems to be a relic of decades past really.Formula sheets give you this information:This means you need to put the equation in the form shown but you may realise it needs to be generalised a bit into:and we need to find a, b and c
How do you find the focus of the parabola x^2+6y=0?
x² + 6y = 0 6y = - x² y = - 1/6 x² Vertex (0, 0): h = 0 k = 0 a = - 1/6 p = 1 / 4a: p = 1 / 4(- 1/6) p = 1 / (- 4/6) p = 1 / (- 2/3) p = - 3/2 p = - 1.5 Focus (Fx, Fy): Since y graphs as a downward-opening parabola, the focus will be p units below the vertex, so Fx = h: Fx = 0 Fy = p - k Fy = - 1.5 - 0 Fy = - 1.5 Focus (0, - 1.5) ¯¯¯¯¯¯¯¯¯¯¯¯¯
Find the vertex, focus and directrix?
1. V: (-3,2) (y-2) = (1/16) (x+3)² a = 1/16 => If D = | F - V | => a = 1/(4D) => D = 4 F = (-3,2+4) = (-3,6) d: y=2-4 = -2 Comprobation: (x+3)² + (y-6)² = (y+2)² => (x+3)² + y²-12y+36 = y²+4y+4 => (x+3)² = 16y-32 = 16(y-2), ok 2. y² + 6y + 12x + 33 = 0 => (y+3)² - 9 = -12x - 33 => (x+2) = -(1/12)(y+3)² V: (-2,-3) D = 3 F = (-2-3,-3) = (-5,-3) d: x = -2+3 = 1 Compr: (x+5)²+(y+3)² = (x-1)² => x²+10x+25+y²+6y+9=x²-2x+1 => 10x+25+y²+6y+9 = -2x+1 => y²+6y+12x+33 = 0, ok 3. y+12x-2x² = 15 => -2(x-3)² + 18 + y = 15 => y+3 = 2(x-3)² V: (3,-3) D = 1/8 F = (3,-3+1/8) = (3,-23/8) d: y=-3-1/8=-25/8 Ellipses: 1. x²/16+y²/12=1 a=4, b=2√3 C = (0,0) c = √(a²-b²) = √4 = 2 F₁ = (0+2,0) = (2,0) F₂ = (0-2,0) = (-2,0) Vertices V₁ = (0+2√3,0) = (2√3,0) V₂ = (0-2√3,0) = (-2√3,0) 2. 16x² - 32x + 9y² = 128 => 16(x²-2x) + 9y² = 128 => 16(x²-2x+1) + 9y² = 128+16 => 16(x-1)² + 9y² = 144 => (x-1)²/3² + y²/4² = 1 a=3, b=4 c = √(16-9) = √7 C=(1,0) F₁ = (1,0+√7) = (1,√7) F₂ = (1,0-√7) = (1,-√7) V₁ = (1,0+4) = (1,4) V₂ = (1,0-4) = (1,-4) 3. x² + 3y² + 6x - 18y + 33 = 0 => (x+3)² - 9 + 3(y-3)² - 27 + 33 = 0 => (x+3)² + 3(y-3)² = 3 => (x+3)²/(√3)² + (y-3)² = 1 a=√3, b=1 c = √2 C = (-3,3) F₁ = (-3+√2,3) F₂ = (-3-√2,3) V₁ = (-3+√3,3) V₂ = (-3-√3,3)
What is the length of the latus rectum of a parabola with focus (3, -4) and directrix x+y+7=0?
Note that the latus rectum of a parabola is a chord that:Passes through the focus, andIs parallel to the directrix and, therefore, has the same slope as the directrix.Let [math]S(3, -4)[/math] be the focus given. Given the directrix with equation[math]\displaystyle x+y+7 = 0,\ldots (1)[/math]let the equation of the latus rectum be[math]\displaystyle x+y+k = 0[/math]Since the focus [math]S(3, -4)[/math] lies on the latus rectum, we must have[math]\displaystyle 3-4+k = 0[/math][math]\displaystyle \therefore k = 1,[/math]and the equation of the latus rectum is[math]\displaystyle x+y+1 = 0. \ldots(2)[/math]We now obtain the equation of the parabola and, solving this along with equation (2), find the endpoints of the latus rectum. The Distance Formula will then yield the desired length.Let us derive the equation of the parabola from first principles. Let [math]P(x, y)[/math] be a point on the locus that is the parabola. Let [math]M[/math] be the foot of the perpendicular drawn from [math]P[/math] on to the directrix. Then [math]PM[/math] is the perpendicular distance of point [math]P(x, y)[/math] from the directrix given by equation (1), i.e.[math]\displaystyle PM = |{\frac{x+y+7}{\sqrt{1^2+1^2}}}| = \lvert{\frac{x+y+7}{\sqrt{2}}}\rvert.[/math]The distance [math]PS[/math] of [math]P(x, y)[/math] from the focus [math]S(3, -4)[/math] is, by the Distance Formula,[math]\displaystyle PS = \sqrt{(x-3)^2+(y+4)^2}[/math]For the locus of point [math]P(x, y)[/math] to trace a parabola, we must have, by the definition of parabola,[math]\displaystyle PS = PM[/math][math]\displaystyle PS^2 = PM^2[/math][math]\displaystyle {(x-3)^2+(y+4)^2} = \frac{(x+y+7)^2}{2}[/math]which, upon reduction, yields the equation of the parabola:[math]\displaystyle x^2+y^2 -26x + 2y - 2xy + 1 = 0 \ldots(3)[/math]From (2),[math]\displaystyle y = -(x+1) \ldots(4)[/math]Substituting in (3), we obtain[math]\displaystyle x^2 + (x+1)^2 –26x - 2(x+1) + 2x(x+1) + 1 = 0[/math][math]\displaystyle \therefore x^2 - 6x = 0[/math][math]\displaystyle \therefore x = 0, 6.[/math]Substituting in (4), we obtain[math]\displaystyle y = -1, -7.[/math]The extremities of the Latus Rectum are, therefore, the points [math]A(0, -1)[/math] and [math]B(6, -7)[/math], and its length [math]AB[/math] is obtained by using the Distance Formula:[math]\displaystyle AB = \sqrt{(6–0)^2+(-7+1)^2}[/math][math]\displaystyle \therefore AB = 6\sqrt{2}[/math] units.
What is the equation of the parabola whose focus is (0,-2) and directrix is y=2?
Let, P (h,k) be a point on the parabola.Focus S(0,-2) and a perpendicular from P cuts the directrix at MSP= √[h²+(k+2)²]PM = | k-2|For parabola, SP= PMor, SP²= PM²or, h²+(k+2)²= (k-2)²or, h²+8k=0The equation of parabola isx²+8y=0or, x²= -8y
Find the vertex and focus point of the parabola 3x^2-6x-6y+10=0?
Convert to standard form, by completing the square: 3(x² - 2x) = 6y-10 3(x² - 2x + 1) = 6y-10+3(1) (x-1)² = ( 6y - 7)/3 (x-1)² = 2y - 7/3 (x-1)² = 2(y - 7/6) Vertex at (1, 7/6) Focal length 4p=2, so p=1/2 Focus at (1, 5/3)
How to find the vertex, focus, and directrix of (x-2)^2=8(y+3)?
First I'll put this in standard form: (y + 3) = (1/8) * (x - 2)^2 The vertex is found as follows: (x - 2) = 0 implies that x = 2. 2 is the x-coordinate of the vertex. (y + 3) = 0 implies that y = -3. -3 is the y-coordinate of the vertex. The vertex is (2, -3).....<<<<<.....Vertex. --------------------------- This parabola opens up because: (1) It is the x-term that is squared, thus it either opens up or down. (2) Since the x^2 term is positive it must open up. Therefore the standard formula (x-2)^2 = 4p*(y+3) applies to this parabola. Since we started with (x-2)^2 = 8 * (y + 3) and these formulas are equivalent to one another: 4p = 8 p = 2 p is a very critical number because it gives the distance between the vertex and the focus and the distance between the vertex and the directrix. Since the parabola opens up, the axis of symmetry is vertical. Since it passes through x = 2 (the vertex) the equation for the axis of symmetry is the equation x = 2. Since the vertex is (2, -3), the focus is p = 2 units above the vertex and it is (2, -1) ---------------------- Since the directrix is perpendicular to the axis of symmetry, its equation is y = C for some constant C. However the y-coordinate of the vertex is (-3) and p = 2 and the directrix is two units below the vertex (it is on the opposite side from the focus): The equation of the directrix is y = -5 --------------------- Summarizing ***************************** The vertex is (2, -3) The focus is (2, -1) The directrix is the equation y = -5 ***************************** .