Q8.
A curve has equation y = kx2 + 1 and a line has equation y = kx, where k is a non-zero
constant.
(0)
Find the set of values of k for which the curve and the line have no common points.
(ii)
State the value of k for which the line is tangent to the curve and, for this case, find
the coordinates of the point where the line touches the curve.
Nov 2010 |V2Q6 7 Marks
Ans: i)0
Answers
- SORRY I DIDN'T KNOW THE ANSWER
Given : A curve has equation y = kx² + 1 and a line has equation y = kx,
k is a non-zero
To Find : the set of values of k for which the curve and the line have no common points.
the value of k for which the line is tangent to the curve and find
the coordinates of the point where the line touches the curve.
Solution:
y = kx² + 1
y = kx
Equate y
kx² + 1 = kx
=> kx² -kx + 1 = 0
Quadratic equation is of the form ax²+bx+c=0 where a , b and c are real also a≠0.
D = b²-4ac is called discriminant.
D >0 roots are real and distinct
D =0 roots are real and equal
D < 0 roots are imaginary ( not real ) and different
For no common solution D < 0
For tangent D = 0
kx² -kx + 1 = 0
D = (-k)² - 4k = k² - 4k = k(k - 4)
k(k - 4) < 0
if 0 < k < 4
Hence values of k for which the curve and the line have no common points.
is k ∈ (0 , 4)
for tangent
k(k - 4) = 0
k is non zero
Hence k = 4
y = 4x² + 1
y = 4x
4x² + 1 = 4x
=> 4x² - 4x + 1 = 0
=> (2x - 1)² = 0
=> x = 1/2
y = 2
Point of tangency = ( 1/2 , 2)
point where the line touches the curve. ( 1/2 , 2)
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