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Tangents and Normals (Part-3)
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This lesson concludes the discussion on tangents and normals

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Unacademy user
mam ye term kaise nikali apne...??
PC
sir hit and trial ka koi hint do na solve karne ka. ye wala part smj ni aaya baaki sb bilkul fit ho gya mind me.
Modassir
a month ago
hidden trial method is very easy Try with smallest real no. to satisfy equation except zero
sir plz plz plz tell me a name of educator who completed or complete the syllabus of physics for IIT JEE till december.complete syllabus plz sir reply krna
Vineet Loomba
a year ago
i am not sure sagar ...otherwise i would have told you ..sorry
Sagar Seth
a year ago
thaks sir for replying.can u complete our physics syllabus because sir i see neither on unacadmy who is compliting physics syllabus for IIT JEE
sir can you suggest me a particular book name for solving more types of problems and which is also good in theory
Vineet Loomba
a year ago
rd sharma is good
thank u sir
Sir, Asking to do a long division in 2nd question, I do not understand which term will be the dividend and which will be divisor????.....
  1. IIT-JEE VIDEO COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) IIT-JEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR Search vineet loomba unacademy" on GOOGLE


  2. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!


  3. APPLICATION OF DERIVATIVES TANGENTS AND NORMALS IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)


  4. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 2 Example Find the equation of all lines having slope 2 and being tangent to the curve y + 2 So Slope of the tangent to the given curve at any point (x, y) is given by - dx (x -3)2 2 But the slope is given to be 2. Therefore 2 (x-3)2 or (x 321 or x 31 or x -2, 4 Now x - 2 gives y 2 and x 4 gives y2. Thus, there are two tangents to the given curve with slope 2 and passing through the points (2, 2) and (4, -2). The equation of tangent through (2, 2) is given by y-2-2(x-2) or y-2x+2-0 and the equation of the tangent through (4, -2) is given by y-(-2) = 2(x-4) or y-2x + 10 = 0 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  5. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The equation of the tangent drawn to the curve y2-2x3-4y + 8 0 from the point (1,2) is (B) y-(2+2/3)--2/3x-2) (D) y-(2-23)--2/BCx-2) (A) y-(2+243)-23x-2) (C)-(2-2/3)-23(r-2) SOLUTION: (AD) 33-3h -4k-4-0 ...(i) Let tangent drawn from (1, 2) to the curve y2-2r3-4y +8-0 meets the curve in point (h, k) Equation of tangents at (h, k) Slope of tangent at (h, k) Adding (i) and (ii), we get (h +1)(h-2)2=0 Forh k is imaginary. So consider only h-2. Using (ii) and h-2, we get dy 3 x drth, k ) y-2](h, k ) k -2 Equatoin oftangent is Thus tangent from (1, 2) meets the curve in points 3h k-2 As tangent passes through (1,2), (2, 2 213)-and (2, 2-2V3). Equation oftangents at these points are y-(2+2V3)-2V3 (x-2) we can obtain and y-(2-2V3)--2V3 (x-2) 3h k -2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  6. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The slope of the tangent of the hyperbola 2x2 - 3y2 6 at (3, 2) is (2) 1 (4) 2 (3) 0 Sol.(2) Differentiating the given equation of the curve 4x - 6y. y - Q dx dy 2x dx 3y dy 2 3 dx 82) 3 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  7. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The tangent to3ay at the point A (at, at) meets the curve against at the points 2 t3 at at at (A) 4 8 (C) (4at2, 8at3)(D) (4ai2,-8at3) (B) 48 SOLUTION: (B) Equation of tangent to : a, y is 3a a(3) 32++t t+ t at 2 at 32+322t2+2t, i.e. 3 tr 2y a3 0 Let B (at,at) be the point where it again meets the curve. Hence the meeting point B is Slope of tangent at A slope ofAB at-at MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  8. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ANGLE OF INTERSECTION OF TWO CURVES If two curves y - f,(x) and y x) intersect at a point P, then the angle between their tangents at P is defined as the angle between these two curves at P. But slopes of tangents at P aand intersection (0) is given by nts at P are( ),ard ( ),.so at p tat arge of . so (dy/dx), -(dy/dx)2 1+(dy/dx)1(dy/dx)2 tan = The other angle of intersection will be (180 - Note : If two curves intersect orthogonally i.e. at right angle then so the condition will be 2 dy) (dy MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  9. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The angle of intersection between the curves y = x and y2= 4x at (4.4) (1) tan-l Ex. (2) tan-1 2 4 2 Sol.(3) Differentiating given equation, we have dy1 and dy .. at (4, 4) dx1 2 Hence angle of intersection tan1 tan (1/3) 1+1 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  10. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The condition that the curves-1 and xy mutually intersect orthogonally is (1) a2 + b2-o (2) a2 b2 (3) a2b20 (4) a2 =-2 Sol.(2) The given curves are and xy c2 from (a) bx (1) and (2) intersect orthogonally if and from (2), dy dy dy a2 b2which is the required condition. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)