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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=kxy=kx touches a parabola y=(x1)2y={{(x-1)}^{2}}, then the values of kk are:
(a) 2,22,-2
(b) 0,40,4
(c) 0,20,-2
(d) 0,20,2
(e) 0,40,-4

Explanation

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 yy 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)y=kx................(i)
Equation of parabola: y=(x1)2...........(ii)y={{\left( x-1 \right)}^{2}}...........(ii)
Now, when line & parabola meets, for the given condition we need to equate equation (i) with equation (ii) because the intercepts of yy in both the curves at the meeting point will be the same.
kx=(x1)2\Rightarrow kx={{(x-1)}^{2}}
As per identity rule (a+b)2=a2+2ab+b2{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} ,we can expand (x1)2{{(x-1)}^{2}}
kx=x22x+1\Rightarrow kx={{x}^{2}}-2x+1
Now we will rearrange the equation for getting all the xx terms at one side,
x22xkx+1=0.........(iii)\Rightarrow {{x}^{2}}-2x-kx+1=0.........(iii)
Now, taking x as common and converting the above equation to quadratic form of ax2+bx+c=0a{{x}^{2}}+bx+c=0
x2(2+k)x+1=0\Rightarrow {{x}^{2}}-(2+k)x+1=0
So here, a=1,b=(k+2),c=1................(iv)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=0D=0
b24ac=0{{b}^{2}}-4ac=0
Substituting the values from (iv) we get
[(k+2)]24(1)(1)=0............(v){{[-(k+2)]}^{2}}-4(1)(1)=0............(v)
[k2]24=0\Rightarrow {{[-k-2]}^{2}}-4=0
k2+4k+44=0\Rightarrow {{k}^{2}}+4k+4-4=0
k2+4k=0\Rightarrow {{k}^{2}}+4k=0
k(k+4)=0\Rightarrow k(k+4)=0
k=0;k=4\Rightarrow 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. b24ac=0{{b}^{2}}-4ac=0 if the sign of the coefficient is not put correctly.
Alternatively, we can rewrite, equation (v) as
[(k+2)]24=0{{[-(k+2)]}^{2}}-4=0
(k+2)2=4\Rightarrow {{(k+2)}^{2}}=4
Taking square roots on both sides
k+2=±2k+2=\pm 2
k=0;k=4\Rightarrow k=0;k=-4
So, the answer in this case is again the same.