Question

find the points on the curve y= x^3-3x^2-4 at which the tangent lines are parallel to the line 4x+y-3=0​

Answer 1

Answer:

Step-by-step explanation:

Any line || to the given line is y = 3x + 9 -k. If this line happens to be a tangent line to the given curve y = 2x^2 - x + 1 then the points of intersection of this line and the curve must coincide . For that , we must have the discriminant of the quadratic (2x^2 - x + 1) = 3x + (9 - k) or 2x^2 -4x + (k - 8) =0, equals 0 . That is ,

(b^2 - 4ac) = 16 - 8(k - 8) = 0 ===> k = 10 and the quadratic 2x^2 - 4x + (k -8) become = 2x^2 - 4x +2 =0 or (x - 1)^2 = 0 ==> x = 1 which in turn implies y = 3x + (9-k) = 3x -1 = 2 . So the required point is (1, 2).

This is a general method to show that a given line is a tangent line to a conic

hope it helps

:)

Answer 2

Answer:

(0,0) and (2, -4)

Step-by-step explanation:

Basic idea : We know that, derivative of any curve gives the slope of it's tangent and if tangent is parallel to any line than their slope is same

Given curve =>   y = x^3 - 3x^2 - 4

                         dy/dx = 3x^2 - 6x - 4 --------(1)

 

Now 3x^2 - 6x - 4 should be parallel to slope of 4x + y - 8 = 0

     slope of 4x + y - 8 = 0 is

                  y = -4x + 8 --------(2)

     on comparing (2) with the standard equation y = mx+c, m = slope

                 we get m = -4

  equating (1) and (2) we get,

        3x^2 - 6x - 4 = -4

        3x^2 - 6x = 0

        3x ( x - 2 ) = 0

        either 3x = 0                                    or x - 2 = 0

        so x = 0 , 2

if x = 0

y = 0 ______ by putting value of x in (1)

also if x = 2

         y = -4 _______ similarly

therefore the required points are

     ( 0 , 0 )      and     ( 2 , -4 )