Question

A straight line has equation y = -2x + k, where k is a constant, and a curve has equation y =
2/x-3
(i) Show that the x-coordinates of any points of intersection of the line and curve are given by the
equation 2x² - (6+k)x + (2 + 3k) = 0.
[1]
(ii) Find the two values of k for which the line is a tangent to the curve.
[3]
The two tangents, given by the values of k found in part (ii), touch the curve at points A and B.
(iii) Find the coordinates of A and B and the equation of the line AB.
[6]

Answer 1

In this answer, I suppose that A,B are the points where the line y=−2x+k is tangent to the curve.

For k=2, 2x2−(6+k)x+3k+2=0⇒2x2−8x+8=0, i.e. 2(x−2)2=0. So, x=2, and y=2x−3=22−3=−2. So, we know that (2,−2) is the point where the line y=−2x+2 is tangent to the curve.

For k=10, 2x2−(6+k)x+3k+2=0⇒2x2−16x+32=0, i.e. 2(x−4)2=0. So, x=4, and y=2x−3=24−3=2. So, we know that (4,2) is the point where the line y=−2x+10 is tangent to the curve.

Hence, the equation of the line AB is

y−(−2)=2−(−2)4−2(x−2)⇒y=2x-6