## Vineet Loomba is teaching live on Unacademy Plus

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

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!!

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)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The equation of tangent at any point of the curve x = at, y = 2at is- (1) x = ty + at2 (3) ty xat Ex (2) ty + x + at2 = 0 (4) ty = x + at3 1 Sol.(3) dy=(dy/dt) 2a dx/dt) 2at t dx .:. equation of the tangent at (x, y) point is MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find points on the curve 25-1 at which the tangent are y-fix) i) parallel to x-axis (ii) parallel to y-axis 90 Sol. Differentiating T+25=1with respect to x, we get dy-25 x 2 25 dx i) Now, the tangent is parallel to the x-axis if the slope of the tangent is zero which gives 25 x 0 This is possible if x = 0. Then-+ 1 for x = 0 gives y-25, i.e., y = 5 Thus, the points at which the tangents are parallel to the x-axis are (0, 5) and (0,-5) (ii) the tangent line is parallel to y axis if the slope of the normal is 0 which give Therefore, = 1 for y-0 gives x = 2. Hence, the points at which the tangents are parallel to the y-axis are (2, 0) and (-2, 0) 4 25 s50, i.e, y 0 4 25 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The equation of the tangent to the curve x2(x - y) a2(x y) 0 at origin is (1) x y+ 10 (3) x + y = 0 (2) x y 2 0 (4) 2x y 0 Sol.(3) The given equation of the curve is Differentiating it w.r.t. x 2 dy +a2.dy = 3x2 + 2xy _ a2 3x2-2xy-x2 dy +a2 + a2 dy Now at origin i.e. x = 0, y = 0 . a (1 dy/dx) 0 dy/dx-1 dx the equation of tangent is y-0--1 (x-0) -y-- : > x + y-0 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The abscissa of the point on the curve ay, the normal at which cuts of equal intercept from the axes is SOLUTION: (D) Since the normal makes equal intercepts on the axes its inclination to axis ofx is either 45 or 135 So two normal are possible with slopes 1 and The given curve is ayr3 Differentiate to get: dy 3x dy dr dx 2ay 2 ay = 1 ai The slope ofnormal One squaring 4ay 9x4. Using (i) we get: MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

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)

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)

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)

FOR 100% SUCCESS IN 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. * Share among your peers as SHARING is CARING!!