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Tangents and Normals Examples (Part-2)
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This lesson further explains the concept of tangents and normals with more examples

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
💥IITian | Top Educator 💥All Maths Chapters Complete & FREE 💥10+ Years Experience 💥Youtube: Maths Wallah 💥DPPs @ vineetloomba.com

U
Unacademy user
yes sir english ka kuch upay batao...
Siddhant Jain
a year ago
Unacademy pe ek Educator h Shalini Bhatt.. Unki videos Dekho
sir i didn't got the last question how can the normal make 45 or 135 degree and can cut the equal intercept from the axes i tried to rewind and get the doubt clear but it got solved
*it didn't got solved
sir isme advance previous year ke questions bhi add kare
DS
please provide course on mean value theorem sir plzz my entrance exam is upcoming sir
Vineet Loomba
10 months ago
I ll add it soon in chapterwise important questions for board and xams
Vineet Loomba
10 months ago
Or do u need it for main
DS
Dipali Singh
10 months ago
i am bca final year student and i need the course for my mca entrance exam , which is held on 27th of april
DS
Dipali Singh
10 months ago
please provide lecture of mean value theorem sir
Nt
sorry sir clear Ho gya
Vineet Loomba
a year ago
good to know
  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) 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)


  5. 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)


  6. 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)


  7. 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)


  8. 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)


  9. 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)


  10. 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)


  11. 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!!