let C b d curve y=x3(x takes all real values) the tangent at A except(0,0) meets the curve again at - Maths - Application of Derivatives

Question

let C b d curve y=x3(x takes all real values) the
tangent at A except(0,0) meets the curve again at B if the gradient
at B is k times the gradient at A then k is equal to?

Anuradha Sharma · Accepted Answer

For solving such types of problem we will assume that there is a
point that lies on the graph and then we will find out the equation
of the tangent at that point and after that we will solve this
tangentwith the given curve to find the point where it will cut
again and then we can find the slope of the curve at that pooint to
find the value of k The given curve is, y = x3So tangent at any
point let say 1,1 that is lying on this curve , can be gicenn
asfirst finding the slope, dydx= 3x2So slope at point
1,1=3×12=3SO equation of tangent at 1,1,y -1=3x-1y = 3x-2Now
solving this with y = x3 we get,3x-2=x3x3-3x+2=0clearly by hit and
trial method checking for factors of constant term 2 that are
±1, ±2 x = 1 and x= -2 are the solutions but we had
already taken x = 1 so x = -2hence y = -23=-8Hence coordinates
other than A1,1 is B-2,8Now gradient at -2,8=3x2=3×4=12And
here it is given that gradient at B is k times gradient at A so, k
= gradient at Bgradient at A=123=4So k = 4 Here we have found out
the value of k at a particular value and it will satisfy for all
other values too because if it is general form then it should
satisfy at all other points too although we have proved the reverse

let C b d curve y=x3(x takes all real values).the tangent at A except(0,0) meets the curve again at B. if the gradient at B is k times the gradient at A then k is equal to?

let C b d curve y=x³(x takes all real values).the tangent at A except(0,0) meets the curve again at B. if the gradient at B is k times the gradient at A then k is equal to?