Focal chord length of parabola
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 length of the focal chord of parabola \( y^{2}=4 a x \)P that makes an angle \( \alpha \) with the \( x \)-axis, is:W.(1) \( 4 a \sec ^{2} \alpha \)(2) \...
Focal chord length of parabola
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WebMar 27, 2024 · Point of intersection in fourth quadrant gives me a ∈ [ 0, 1) So, parabola is y 2 = 4 ( a 2 − a + 1) x + 5 I know that length of focal chord is a ( t + 1 t) 2 for y 2 = 4 a x with end end of focal chord being ( a t 2, 2 a t) Also, if the focal chord makes angle θ with x-axis then length of focal chord is 4 a csc 2 θ WebApr 6, 2024 · Length of focal chord c = 4 a 3 P 2. Hence, we got the required length as 4 a 3 P 2. Note: The length of a focal chord of a parabola varies inversely as the square of the distance from its vertex. If …
WebFeb 3, 2024 · If a chord is drawn parallel to that focal chord which passes through vertex of parabola at (0,0) , it's length comes out to be $4acosec^2\theta cos\theta$, it's quite easy to prove this using parametric coordinates for the parabola , I'm looking for an intuitive geometric demonstration that AB=A′B′.The equality certainly holds but I feel ...
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 + … WebFOCAL CHORD : A chord of the parabola, which passes through the focus is called a FOCAL CHORD. ... Also prove that CG = e2CN, where PN is the ordinate of P. x 2 y2 Q.16 Prove that the length of the focal chord of the ellipse 1 which is inclined to the major axis at a 2 b2 2ab 2 angle is . a 2 sin 2 b 2 cos2 ...
WebAfter the properties of a parabola, let’s study the focal chord. The chord which passes through the focus is called the focal chord of the parabola. The focal distance of some …
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 charles mardon net worthWebThe length of a focal chord of the parabola y 2=4ax at a distance b from the vertex is c. Then A 2a 2=bc B a 3=b 2c C ac=b 2 D b 2c=4a 3 Hard Solution Verified by Toppr Correct option is D) Equation of the focal line passing through (a,0) is y=m(x−a) The distance of this line from the vertex is b. ⇒b= ∣∣∣∣∣ 1+m 2am ∣∣∣∣∣ ⇒b 2(1+m 2)=a 2m 2 .... (1) harry potter wand collection bookWebAssertion A: The least length of the focal chord of y 2 = 4 a x is 4 a. Reason R: Length of the focal chord of y 2 = 4 a x which makes an angle θ with its axis is 4 a cosec 2 θ . Medium harry potter wand cookiesWebLength 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 harry potter wand cost galleonsWebAssertion A: The least length of the focal chord of y 2 = 4 a x is 4 a. Reason R: Length of the focal chord of y 2 = 4 a x which makes an angle θ with its axis is 4 a cosec 2 θ. harry potter wand collection for saleWebThe 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 … charles mares obituaryWebThe length of the intercept on the normal at the point (a t 2, 2 a t) of the parabola y 2 = 4 a x made by the circle which is described on the focal distance of the given point as diameter is. Hard. View solution > If the tangent and normals at the extremities of a focal chord of a parabola intersect at (x 1 ... charles mares hardware new carlisle in