WebSolution : Let the equation of the hyperbola be x2 a2 − y2 b2 = 1 x 2 a 2 − y 2 b 2 = 1 and the coordinates of P be ( h, k ). Any tangent of slope m to this hyperbola will have the … WebIf the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B , then the locus of the mid point of AB is : 2400 71 JEE Main JEE Main 2024 Application of Derivatives Report Error A x2 − 4y2 +16x2y2 = 0 B x2 − 4y2 −16x2y2 = 0 C 4x2 −y2 +16x2y2 = 0 D 4x2 −y2 −16x2y2 = 0 Solution: 4y2 = x2 +1
Tangents are drawn to the hyperbola 4x^2 − y^2 = 36 at …
WebCorrect option is C) given hyperbola is: 4x 2−9y 2=36 let (x 0,y 0) be point of contact of normal on the hyperbola Finding slope of normal at that point: Differntiating hyperbola equation we get; 4×2×x−9×2×y dxdy=0 ⇒ dxdy= 9y4x = slope of tangent ∴ slope of normal= 4x−9y equation of normal at (x 0,y 0) is: y−y 0= 4x 0−9y 0(x−x 0) WebTangents are drawn to the hyperbola 4x2−y2 =36 at the points P and Q. If these tangents intersect at the point T (0,3), then the area (in sq.units) of ΔP T Q is A 36√5 B 45√5 C 54√3 … red hot chili peppers tour miami
Tangents are drawn to the hyperbola 4x^2 - y^2 = 36 at the points …
WebApr 8, 2024 · Hyperbola Answer Tangents are drawn to the hyperbola $4{x^2} - {y^2} = 36$ at the point P and Q. If these tangents intersect at the point$T\left( {0,3} \right)$. Then … WebHyperbola: 4x²-y²=36 or x²/9 -y²/36 = 1. It’s tangent may be written as: ax/9 -by/36 = 1 and it is given that it passes through (0,3). Then, 0 - 3b/36=1 or b=- 12. Substitute this in our … WebTangents are drawn to the hyperbola 4x^2 - y^2 = 36 at the points P and Q. If these tangents intersect at the point T (0, 3) then the area (in sq. units) of Δ PTQ is Question 32 Tangents … red hot chili peppers tour opening act