## 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) SLOPES OF TANGENT & NORMAL: The slope of the tangent to a continuous curve y -fix) at a point Px,, y,) is equal to (, yi) Thus, if the tangent to the curve y f(x) at point (x1, y) makes an angle with x-axis. then tan - f(x) If the tangent at P is parallel to x-axis, then dy dx 0 If the tangent at P is parallel to y-axis, then 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) dy dx (x y1) If the tangent at P is parallel to x-axis, then 0 If the tangent at P is parallel to y-axis, then dy 2 Normal to a curve at a point Px, y is a line perpendicular to the tangent at P and passing through P. Therefore, dx Slope of the normal at P =- Slope of the tangent at P dy dx MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) EQUATIONS OF TANGENT AND NORMAL Equation of a straight line passing through (x, y,) and having slope m is y - y,- m(x-x) Tangent and normal to the curve y = f(x) at point (x1, y) pass through P and have slopes respectively dy dx (xV) (x,,y Equation of tangent to y - fx) at (x, y,) is; Equation of the normal to y = f(x) at (x1, y) is : (x -x1) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) I dy Note-1 : lf-) = , then the tangent at P(x1, y) is parallel to y-axis and its equation is x = x1. Note-2 If y-0, then the normal at Px, y,) is parallel to y-axis and its equation is y y Note-3 : The equations of tangent and nomal to the curve having its parametric equations x- ft) and y -gt) are given by g y-g(t) = (x-ft)) Equation of tangent) -f(t) g (t) and y -gix(Equation of nomal) respectively. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: Find the point at which the tangent to the 2 curve y4x-3 -1 has its slope Sol. Slope of tangent to the given curve at (x, y) is dy 1 dx 2 2 4x-3 2 2 2 The slope is given to be .So 4x-3 3 or 4x -3 -9 or x-3 Nowy-\4x-3 -1. So when x 3, y-4(3)-3 Therefore, the required point is (3, 2) 1-2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The equation of tangent to the curve y2= 6x at (2,-3) (1)x+y 1-0 (3) x-y+1=0 Ex (2)x +y 10 (4) x y2 0 Sol. (2) Differentiating equation of the curve with respect to x dy 2y6 3 dy dx2.-3) Therefore equation of tangent is y+3(x - 2)xy10 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

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)