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Equation of directrix of parabola y2+4y+4x+2

WebStep 1: Standardize the given equation. The equation y 2 + 4 x + 4 y + k = 0 of the parabola is given. Standardize the given equation as follows: Add and subtract 4 from the Left-hand side. y 2 + 4 x + 4 y + k + 4 - 4 = 0 ⇒ y 2 + 4 y + 4 + 4 x + k - 4 = 0 ⇒ y + 2 2 = 4 - 4 x - k ⇒ y + 2 2 = - 4 x + - 4 + k 4 ... 1 WebGiven the focus (h,k) and the directrix y=mx+b, the equation for a parabola is (y - mx - b)^2 / (m^2 +1) = (x - h)^2 + (y - k)^2. Equivalently, you could put it in general form: x^2 + …

Focus and Directrix of a Parabola - GeeksforGeeks

WebNov 24, 2024 · The standard equations of the parabola with the given coordinates of vertices, foci and equation of directrix are as follows: Vertex: (0, 0) Focus: (a, 0) Directrix: x = -a Equation of parabola: y 2 = 4ax Parametric Coordinates: (at2, 2at) Vertex: (0, 0) Focus: (-a, 0) Directrix: x = +a Equation of parabola: y 2 = -4ax WebWe start by assuming a general point on the parabola (x,y) (x,y). Using the distance formula, we find that the distance between (x,y) (x,y) and the focus (-2,5) (−2,5) is \sqrt { … schematherapeut junior https://lcfyb.com

How do you find the focus, vertex, and directrix of #y^2=-4x#?

WebQuestion: QUESTION 1 Find the focus and directrix of the parabola with the given equation. y^(2)=4x. QUESTION 1 Find the focus and directrix of the parabola with the given equation. y^(2)=4x. Expert Answer. Who are the experts? Experts are tested by Chegg as specialists in their subject area. We reviewed their content and use your … WebFind the focus, directrix, and focal diameter of the parabola. y2 = 8x. write the equations of the parabola, the directrix, and the axis of symmetry. vertex: (-4,2) focus: (-4,6) if … WebA parabola with focus (3, 0) and directrix x = –3. Points P and Q lie on the parabola and their ordinates are in the ratio 3 : 1. The point of intersection of tangents drawn at points P and Q lies on the parabola (1) y 2 = 16x (2) y 2 = 4x (3) y 2 = 8x (4) x 2 = 4y schematherapeut helmond

Conic Section (Para Ellip Hyper) PDF Ellipse Perpendicular

Category:A parabola with focus (3, 0) and directrix x = –3. Points P and Q …

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Equation of directrix of parabola y2+4y+4x+2

The equation y^2 + 4x + 4y + k = 0 represents a parabola …

WebGiven the focus (h,k) and the directrix y=mx+b, the equation for a parabola is (y - mx - b)^2 / (m^2 +1) = (x - h)^2 + (y - k)^2. Equivalently, you could put it in general form: x^2 + 2mxy + m^2 y^2 -2 [h (m^2 - 1) +mb]x -2 [k (m^2 + 1)^2 -b]y + (h^2 + k^2) (m^2 + 1) - b^2 = 0 At least, I think this last one is right. WebThe parabola will be upward facing, with the vertex at the point midway between the focus and the directrix, so its vertex will be at (-8, -1). The distance from the focus to the …

Equation of directrix of parabola y2+4y+4x+2

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WebJan 20, 2024 · Explanation: Write y2 + 4x −4y − 8 = 0 in the form of equation [1]: x(y) = − 1 4y2 + y + 2 [2] Matching values from equation [2] with variables in equation [1]: a = − 1 … WebGiven an equation of a parabola, find the vertex, focus, and directrix. Draw the graph y 2 -4y+4=4x+4 Expert Answer 1st step All steps Final answer Step 1/3 Given is the …

WebMar 28, 2024 · Equation of the directrix: y = -a Equation of axis: x = 0 Length of the latus rectum: 4a Focal distance of a point P (x, y): a + y 4. x2 = – 4ay Here, Coordinates of vertex: (0, 0) Coordinates of focus: (0, -a) Equation of the directrix: y = a Equation of axis: x = 0 Length of the latus rectum: 4a Focal distance of a point P (x, y): a – y WebWrite the equation into the standard form of the equation of the parabola: x2 + 4x – 4y = 0.… A: Click to see the answer Q: Find the focus and directrix of the parabola with the given equation. 4) 9x2 = -5y A: Compare standard equation of parabola question_answer question_answer question_answer question_answer question_answer question_answer

Weby2 = 4x y 2 = 4 x Rewrite the equation in vertex form. Tap for more steps... x = 1 4y2 x = 1 4 y 2 Use the vertex form, x = a(y−k)2 +h x = a ( y - k) 2 + h, to determine the values of a a, h h, and k k. a = 1 4 a = 1 4 h = 0 h = 0 k = 0 k = 0 Since the value of a a is positive, the parabola opens right. Opens Right Find the vertex (h,k) ( h, k). WebThe given equation, y²+4y − 4x − 2 = 0 can be written as ; (y+2)² = 4 + 4x +2 = 4 (x+3/2) or Y² = 4aX , where Y = (y+2) , X = (x+3/2) and 4a = 4 (or a = 1 ) . Now, vertex V (x, y) is given by X= 0 = Y ==> (x, y) = (-3/2, -2) . Focus is given by F (x, y) i.e. X = a = 1 & Y = 0 ==> (x, y) = (-1/2, -2) and lastly the rectum = 4a = 4 . Gary Ward

WebJan 26, 2016 · The focus is at (1,3), the vertex is (2,3) and the directrix is at x = 3 Explanation: Reformulate the equation to have one variable on its own on the left hand side. In this case it should be x because y is the one that is raised to a power. y2 + 4x − 4y −8 = 0 4x = −y2 + 4y +8 x = − 1 4(y2 −4y − 8)

WebNov 21, 2015 · Find the equation of the directrix of the parabola y 2 + 4 y + 4 x + 2 = 0 I tried it as follows: ( y + 2) 2 + 4 x − 2 = 0 ( y + 2) 2 = − 4 ( x − 1 2) On comparing with Y 2 = − 4 a X, a > 0, I got a = 1 Thus, the equation of the directrix is X = a ⇒ X − 1 = 0 . Since the origin has been shifted, the equation is x − 1 2 − 1 = 0 2 x − 3 schematheque hitachiWebJan 20, 2024 · The equation of the tangent at the point P (t), where t is any parameter, to the parabola y2 = 4ax is: Q4. For the curve x2y2 = a2 (x2 + y2), the asymptotes parallel to the coordinates axes are Q5. An equilateral triangle is inscribed in a parabola x2 = 3 y where one vertex of the triangle is at the vertex of the parabola. schematherapie arkinWebNov 21, 2015 · Find the equation of the directrix of the parabola y 2 + 4 y + 4 x + 2 = 0. I tried it as follows: ( y + 2) 2 + 4 x − 2 = 0. ( y + 2) 2 = − 4 ( x − 1 2) On comparing with Y … schematherapie congres 2021WebNov 26, 2016 · y2 = − 4x It is in the form y2 = − 4ax If it is so, then - Its vertex is (0,0) Its focus is ( − a,0) Its directrix is x = a Apply this in the given equation For better understanding the given equation can be written as y2 = 4 ⋅ ( −1) ⋅ x Then- Its vertex is (0,0) Its focus is ( − 1,0) Its directrix is x = 1 Answer link schematherapie coconWebFeb 5, 2024 · x − 2 y + 4 = 0 and x − 3 y + 9 The point of intersections with the parabola y 2 = 4 x were found out to be ( 4, 4) and ( 9, 6) Let R be ( 9, 6). Hence circle C 2 passes through (9,6) and focus (1,0) This data isn’t enough to find the radius of the circle. How do I get more information? conic-sections Share Cite Follow asked Feb 5, 2024 at 11:40 schematherapeut stuttgartWebThen graph the parabola. y^2 = 4x; Find the vertex, focus, directrix, and axis of symmetry of the parabola (y - 1)^2 = 16x. Find directrix focus and axis for the parabola y^2 + 8x - 6y + 1 = 0. Find the focus and directrix of the parabola given by the equation y = 2 x^2 - 3 x + 10. Find the equation of a parabola with directrix x = 2 and focus ... schematherapie boek pdfWebAny point (𝑥, 𝑦) on the parabola is equidistant to the focus and the directrix. We can express these distances using the distance formula, and we get √ ( (𝑥 − 9)² + (𝑦 − 0)²) = √ ( (𝑥 − 𝑥)² + (𝑦 − (−4))²) Simplifying and squaring both sides gives us (𝑥 − 9)² + 𝑦² = (𝑦 + 4)² Expanding the … schematherapie cluster c