Focal chord of y 2 16x is a tangent
WebThe focal chord of the parabola (y−2) 2=16(x−1) is a tangent to the circle x 2+y 2−14x−4y+51=0, then the focal chord can be A 0 B 1 C 2 D 3 Medium Solution Verified by Toppr Correct option is B) Was this answer helpful? 0 0 Similar questions If points (au 2,2au) and (av 2,2av) are extremities of the focal chord of a parabola y 2=4ax, then Hard WebGet an expert solution to The focal chord of the parabola ( y − 2 ) 2 = 16 ( x − 1 ) is a tangent to the circle x 2 + y 2 − 14 x − 4 y + 51 = 0 , then slope of the focal chord can be
Focal chord of y 2 16x is a tangent
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WebStep-1 Length of tangent : Given: The focal chord to y 2 = 16 x is tangent to (x – 6) 2 + y 2 = 2. The standard equation of the parabola is: y 2 = 4 a x. Comparing the given … WebMar 14, 2024 · It is given that the focal chord is tangent to the circle which means that the distance of the focal chord from the center of the circle is equal to the radius of the circle. Therefore, we get m x − y − 4 m 1 + m 2 = 2 Now we will put the value of x = 6 and y = 0 in the above equation, we get ⇒ 6 m − 0 − 4 m 1 + m 2 = 2
WebQ.3 Find the equations of the tangents to the parabola y2 = 16x, which are parallel ... y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2 Q.10(a) The slope of the focal chords of the parabola y2 = 16x which are tangents to the circle (x ... [ JEE 2003 (Scr.)] Q.6 The line 2x + 6 y = 2 is a tangent to the curve x2 – 2y2 = 4. The point ... WebHere, the focal chord of y 2 = 16 x is tangent to circle (x − 6) 2 + y 2 = 2 ⇒ Focus of parabola as (a, 0) i.e. (4, 0) Now, tangents are drawn from (4, 0) to (x − 6) 2 + y 2 = 2. Since, P A is tangent to circle. ∴ t a n θ = slope of tangent = A C A P = √ 2 √ 2 = 1, or B C B P = − 1. ∴ Slope of focal chord as tangent to circle ...
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 WebThe focal chord of \( y^{2}=16 x \) is tangent to\( \mathrm{P} \) \( (x-6)^{2}+y^{2}=2 \). Then the possible values of theW slope of this chord are:(1) \( 1,...
WebMay 20, 2024 · The equation of common tangent to the curves y^2 = 16x and xy = –4, is : ... If one end of a focal chord of the parabola, y^2 = 16x is at (1, 4), then the length of this focal chord is : asked May 18, 2024 in Mathematics by Jagan (21.2k points) jee mains 2024; 0 votes. 1 answer.
WebSOLUTION. Here, the focal chord of y2 =16x is tangent to circle (x−6)2+y2 = 2. ⇒ Focus of parabola as (a,0) i.e. (4,0) Now, tangents are drawn from (4,0) to (x−6)2+y2 = 2. Since, P … cspan andrew jacksonWebT is a point on the tangent to a parabola y 2 = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then. A. SL = 2 (TN) B. 3 (SL) = 2 (TN) C. ... Let PSQ be the focal chord of … ealing council mashWebOct 14, 2024 · The focal chord to y^2 =16x is tangent to (x–6)^2 +y2 = 2, then the possible value of the slope of this chord, are asked Oct 22, 2024 in Parabola by deepikaben ( … ealing councillors email addressesWebThe equation of a common tangent to the curves, y 2=16x and xy=−4 is A x+y+4=0 B x−2y+16=0 C 2x−y+2=0 D x−y+4=0 Medium Solution Verified by Toppr Correct option is D) Step 1: Use slope form of tangent equation of parabola Equation of tangent to parabola y 2=4ax in terms of slope ’m’ is y=mx+ ma ealing council local mpWebClick here👆to get an answer to your question ️ The focal chord to y ^ 2 = 16 x is tangent to ( x - 6 ) ^ 2 + y ^ 2 = 2 then the possible values of the slope of this chord are Solve Study Textbooks Guides ealing council main switchboardWebA chord which passes through the focus of a parabola is called a focal chord. A given chord will be a focal chord if the point \((0,a)\) lies on it. Substituting these coordinates into the equation of the chord above we have ... and substituting \(x=2ap\). In either case, the gradient of the tangent to \(x^2=4ay\) at the point \(P(2ap,ap^2 ... ealing council missed collectioncspan3 xfinity