Question
Question: If a line \[y=kx\] touches a parabola \(y={{(x-1)}^{2}}\), then the values of \(k\) are: (a) \(2,-...
If a line y=kx touches a parabola y=(x−1)2, then the values of k are:
(a) 2,−2
(b) 0,4
(c) 0,−2
(d) 0,2
(e) 0,−4
Solution
Hint: When two curves meet each other at a point the x and y intercept at that point is the same. So, we will equate y from both the equations of the line and parabola. When two curves meet each other at a point then the quadratic equation satisfying that condition always has real roots.
Complete step by step answer:
Equation of line: y=kx................(i)
Equation of parabola: y=(x−1)2...........(ii)
Now, when line & parabola meets, for the given condition we need to equate equation (i) with equation (ii) because the intercepts of y in both the curves at the meeting point will be the same.
⇒kx=(x−1)2
As per identity rule (a+b)2=a2+2ab+b2 ,we can expand (x−1)2
⇒kx=x2−2x+1
Now we will rearrange the equation for getting all the x terms at one side,
⇒x2−2x−kx+1=0.........(iii)
Now, taking x as common and converting the above equation to quadratic form of ax2+bx+c=0
⇒x2−(2+k)x+1=0
So here, a=1,b=−(k+2),c=1................(iv)
As, the line is touching the parabola; the roots of quadratic equation above must be real, which means,
Determinant of the quadratic equation (iii) must be zero, D=0
b2−4ac=0
Substituting the values from (iv) we get
[−(k+2)]2−4(1)(1)=0............(v)
⇒[−k−2]2−4=0
⇒k2+4k+4−4=0
⇒k2+4k=0
⇒k(k+4)=0
⇒k=0;k=−4
Hence, the final answer is option (e).
Note: There is a possibility of mistake when substituting the values of b in the determinant. b2−4ac=0 if the sign of the coefficient is not put correctly.
Alternatively, we can rewrite, equation (v) as
[−(k+2)]2−4=0
⇒(k+2)2=4
Taking square roots on both sides
k+2=±2
⇒k=0;k=−4
So, the answer in this case is again the same.