Focal chord length of parabola
WebLength of the focal chords of the parabola y 2=4ax at a distance p from the vertex is A p2a 2 B p 2a 2 C p 24a 3 D ap 2 Hard Solution Verified by Toppr Correct option is C) y 2=4ax Slope of OP= Slope of OQ ⇒t 2= t 1−1 ∴ P(at 2,2at) & Q(t 2a, t−2a) Let length of focal chord be C. ∴ (at 2− t 2a)2+(2at+ t2a)2=C ⇒ a 2(t 2− t 21)2+(2a) 2(t+ t1)2=C WebAnswer (1 of 4): For any function y = f(x), between x = x1 and x = x2, the formula for the chord length is integral (x = x1 → x2) sqrt[1 + (dy/dx)^2] dx So if the parabola is given by y = ax^2 + bx + c then dy/dx = 2ax + b (dy/dx)^2 = (2ax + …
Focal chord length of parabola
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WebNov 20, 2013 · This is the length of the focal chord (the "width" of a parabola at focal level). Let x 2 = 4 p y be a parabola. Then F ( 0, p) is the focus. Consider the line that passes through the focus and parallel to the directrix. Let A and A ′ be the intersections of the line and the parabola. Then A ( − 2 p, p), A ′ ( 2 p, p), and A A ′ = 4 p. Share Cite Web(v) Length of the focal chord having t 1 and t 2 as end points is a (t 1 — t 1) 2. (vi) Chord of contact drawn from a point (x 1, y 1) to the parabola y 2 = 4ax is yy 1, = 2a (x + x 1) (vii) Equation of the chord of the parabola y 2 = 4ax, which is bisected at (x 1 , y 1) is given by T = S 1 i.e. , yy 1 — 2a (x + x 1) = y 12 – 4ax
WebAnswer: Consider the parabola: The distance between the vertex and the focus, measured along the axis of symmetry, is the "focal length". The "latus rectum" is the chord of the parabola which is parallel to the directrix and passes through the focus. In fact the “latus rectum” used to be calle... WebThe minimum length for any focal chord is evidently obtained when t =±1, t = ± 1, which gives us the LR. Thus, the smallest focal chord in any parabola is its LR. Example – 8. Prove that the circle described on any focal chord …
WebMar 14, 2024 · Consider a parabola y 2 = 4 a x , parameterize it as x = a t 2 and y = 2 a t, then it is found that if we have a line segment passing through focus, with each points having value of t as t 1 and t 2 for the parameterization, then it must be that: t 1 ⋅ t 2 = − 1 Hope for hints. conic-sections Share Cite Follow edited Mar 14, 2024 at 15:05
WebApr 11, 2024 · We are given a parabola \[{y^2} = 4ax\] Let us assume that the chord cuts the X-axis at point D(a,0) Then according to the question we are given the shortest distance from center to the chord is b. Length of the focal chord is c. The distance \[OD = a\]. Let us assume the focal chord makes an angle x with the X-axis.
WebThe latus rectum of a parabola is the chord that is passing through the focus of the parabola and is perpendicular to the axis of the parabola. The latus rectum of parabola can also be understood as the focal chord which is parallel to the directrix of parabola.The length of latus rectum for a standard equation of a parabola y 2 = 4ax is equal to LL' = 4a. soft-tech wipesWebDec 8, 2024 · Question 4 :$$ $$ Let PQ be a focal chord of a parabola with origin as a focus . Coordinates of point P and Q be (-2,0) and (4,0) respectively . Find length of latus rectum and equation of tangent at vertex of parabola. soft teddy bear rugWebPARABOLA ASSIGNMENT - Read online for free. Scribd is the world's largest social reading and publishing site. PARABOLA ASSIGNMENT. Uploaded by mynameis 1609. 0 ratings 0% found this document useful (0 votes) 0 views. 19 pages. Document Information click to expand document information. slow cooker spiced nutsWebSolution The correct option is A (8, –8) For the parabola y2 = 8x; focus S (2, 0). Given point is P (1 2,2) Slope of ←→ SP is 2−0 1 2−2 = −4 3 Equation to ←→ SP is4x+3y−8= 0 4x+3y−8= 0⇒ 4x=8−3y Substituting this value of 4x in y2 = 8x we get y2 = 2(8−3y) ⇒y2+6y−16−16 =0 ⇒(y+8)(y−2) = 0 ⇒ y= 2or−8 y =−8 ⇒4x =8−3(−8)= 32⇒ x= 8 ∴ point … slow cookers perthWebThe length of a focal chord of the parabola y 2=4ax at a distance b from the vertex is c. Then. A a 2=bc B a 3=b 2c C b 2=ac D b 2c=4a 3 Medium Solution Verified by Toppr Correct option is D) Parabola P:y²=4ax−−(1) Vertex =O(0,0) Focus: F(a,0) Let the Focal chord L be (y−0)=m(x−a) So y=mx−ma−−(2)\ Given b = Distance of O from L. soft teddy bear ukWebThe latus rectum is a focal chord which can be used to find the equation of the parabola. The length of the latus rectum is 4a units, which is useful to form the equation of parabola y2 = 4ax y 2 = 4 a x. What Is The Difference Between … slow cooker spiced mixed nuts recipeWebApr 6, 2024 · Substitute the value you get in the expression of length of focal chord ‘c’ and get the value of c. Complete step-by-step answer: We have been given the equation of parabola as ${{y}^{2}}=4ax$ . We need to find the focal chord of the parabola at a distance p from the vertex. Let us take 2 points on the parabola as P and Q. softtech systems review