## Vineet Loomba is teaching live on Unacademy Plus

IIT-JEE CRASH COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Search vineet loomba unacademy" on GOOGLE IIT-JEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010. Doubts/Feedback/Ideas 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 !!

PARABOLA (CRASH COURSE) FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) # TANGENT TO A PARABOLA IIT-JEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) LINE AND PARABOLA Tangent Secant The line y = mx + c meets the parabola y2 = 4ax in two points real, coincident or imaginary according as a >=< cm condition of tangency is, c = MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) METHOD Let the parabola be y 4ax and the given line be y mx + c Eliminatingy from (i) and (i), then This equation, being quadratic inx, gives two values ofx. Shows that every straight line will cut the parabola in two points may be real, coincident or imaginary according as Discriminant of(W) >, < 0 Hence the line y mx +will always be a tangent to the parabola y 4ax MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) EqUATION OF TANGENT IN VARIOUS FORMS (a Poit form Equation of tangent to the given parabola at its point (x, y1) is yy, = 2a (x + x.) (b) Slope form: Equation of tangent to the given parabola whose slope is 'm', is a 2a y=mx +-, (m to Point of contact is (c) Parametric form: Equation of tangent to the given parabola at its point P(t), is ty = x + at? Note : Point of intersection of the tangents at the point t, & t2 is I at,t2, a( 2) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) SOlvED EXAMPLE Example: A tangent to the parabola y2 = 8x makes an angle of 45 with the straight line Solution Let the slope of the tangent be m y3x5. Find its equation and its point of contact. (JEE MAIn) 3-m = 3m .. tan45 1+3m=t(3-m) m=-2 or 2 As we know that equation of tangent of slope m to the parabola y2 = 4ax is y = mx + and point of contact is for m--2, equation of tangent is y-2x 1 and point of contact is for m equation of tangent is y+ 4 and point of contact is (8, 8) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) SOLVED EXAMPLE Example: Find the equation of the tangents to the parabola y Solution : Equation of tangent to parabola y2 = 9x is 9x which go through the point (4, 10). (JEE MAIN) y=mx + Since it passes through (4, 10) 4m 10 = 4m + => 16m2-40 m + 9 = 0 4m 1 9 m- , 4'4 equation of tangent's are 9 y=-+9 & y=-x+1 4 4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) SOLvED EXAMPLE 4ax (JEE MAIN) Example: Find the locus of the point P from which tangents are drawn to the parabola y having slopes m1 and m2 such that ml + m2 = (constant) Equation of tangent to y2-4ax is y = mx + a/m Let it passes through Ph, k) Solution : m2h-mk + a = 0 m1 +m2 = h - locus of P(h, k) is y2-2ax-AX2 2._ = MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

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. Share among your peers as SHARING is CARING !!