how do you show that line 3x+y-2=0 is a tangent of the curve y=(4x-3)(x-2) and how to find the point of contact??
The first equation, which is a straight line, intersects the second equation, which is a parabola, in 3 ways ...
- it misses the curve, no solution
- it cuts the curve in 2 distinct points, 2 solutions
- it touches the curve in 1 point, one solution
It is the last case we want.
To have a single solution the quadratic must factor into two equal factors of the form (qx+b)^2
first equation: y = 2 - 3x
plug into second
(4x-3)(x-2) = 2-3x
4x^2 -8x + 4 = 0
x^2 - 2x + 1 = 0
(x-1)^2 = 0
one solution ..... x = 1
sub back into y = 2-3x
y = 2-3 = -1
the point is (1,-1)
(check my arithmetic)
To show that a line is a tangent to a curve, we need to prove two conditions:
1. The line intersects the curve at only one point.
2. The slope of the line is equal to the slope of the curve at the point of contact.
Let's solve this step by step:
Step 1: Find the point of intersection
- To find the point(s) of intersection between the line and the curve, we need to solve the system of equations formed by equating both equations:
3x + y - 2 = 0 (Equation of the line)
y = (4x - 3)(x - 2) (Equation of the curve)
- Substitute the value of y from the line equation into the curve equation:
3x + (4x - 3)(x - 2) - 2 = 0
Simplify by expanding and collecting like terms:
3x + 4x^2 - 8x - 6x + 6 - 2 = 0
Combine like terms:
4x^2 - 11x + 4 = 0
- Solve the above quadratic equation to find the x-coordinate(s) of the point(s) of intersection using factoring, quadratic formula, or any other appropriate method.
Step 2: Prove the line is a tangent
- After finding the x-coordinate(s) of the point(s) of intersection, substitute it/them back into either the line equation or the curve equation to determine the y-coordinate(s) of the point(s) of intersection.
- Calculate the derivative of the curve equation y = (4x - 3)(x - 2) to find the equation for the slope(s) of the curve at the point(s) of intersection.
- Calculate the slope of the line 3x + y - 2 = 0 by rearranging it into the form y = mx + c, where m represents the slope.
- Compare the slope(s) of the curve to the slope of the line at the point(s) of intersection. If they are equal, the line is a tangent.
The point(s) of contact between the line and the curve will be the one(s) obtained from Step 1.