WebA straight line parallel to y=root3 x passes through q(2,3) and cuts the line 2x+4y-27=0 at P. Find the length of PQ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A straight line parallel to y=root3 x passes through q(2,3) and ... WebGraph y = square root of 3x. y = √3x y = 3 x. Find the domain for y = √3x y = 3 x so that a list of x x values can be picked to find a list of points, which will help graphing the radical. …
If the line y =√3 x + k touches the circle x 2+ y 2=16, then find the ...
Web6 nov. 2024 · Sol: The line y = √3 x can be written as x = r/2 , y = √3 r/2 If this line cuts the given curve, then r 4 16 + a r 3 3 8 + b r 2 3 4 + c r 2 + d r 3 2 + 6 = 0 Therefore OA. OB.OC.OD = r 1 r 2 r 3 r 4 = r 1 r 2 r 3 r 4 = 96 Let (a, b) and ( λ , μ) be two point on the curve y = f (x). If the slope of the tangent to the… Web13 jun. 2024 · y = – √3x – 2 This is the slope intercept form of the given line. ∴ The slope = – √3 and y – intercept = -2 (ii) Given: √3x + y + 2 = 0 √3x + y = -2 Divide both sides by -2, we get √3x/-2 + y/-2 = 1 ∴ The intercept form of the given line. Here, x – intercept = – 2/√3 and y – intercept = -2 (iii) Given: √3x + y + 2 = 0 -√3x – y = 2 rock n brews lax terminal 1
A straight line parallel to y=root3 x passes through q(2,3) and …
Web22 mrt. 2024 · Question 34 (Choice 1) Find the area of the region bounded by the curves 𝑥^2+𝑦^2=4, 𝑦=√3 𝑥 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 Given Equation of Circle 𝑥2+𝑦2=4 𝑥2+𝑦2=2^2 So, Radius = 2 ∴ Point A (2, 0) and B is (0, 2) Let point where line and circle intersect be point M Required Ar WebIf the line y - √3x+ 3 = 0 cuts the parabola y2 = x + 2 at A and B, then PA.PB is equal to (where P is √3 , 0) Tardigrade. Question. Mathematics. Q. If the line y − 3x + 3 = 0 cuts … Web5 sep. 2024 · Solution : line parallel to y = √3x so, slope of line is √3 = tan60° Line passing through Q (2,3) and cuts 2x + 4y - 27 = 0 at P. Let distance between P and Q = r so parametric equation of line (2 + rcos60° , 3 + rsin60°) = (2 + r/2, 3 + √3r/2) This point should satisfy the equation of line 2x + 4y - 27 = 0 ⇒2 (2 + r/2) + 4 (3 + √3r/2) - 27 = 0 other words for they have