Courses for Kids
Free study material
Offline Centres
Store Icon

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) $0,-4$

Last updated date: 17th Jun 2024
Total views: 52.8k
Views today: 1.52k
52.8k+ views
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: $y={{\left( x-1 \right)}^{2}}...........(ii)$
Now, when line & parabola meets, for the given condition we need to equate equation (i) with equation (ii) because the intercepts of $y$ in both the curves at the meeting point will be the same.
$\Rightarrow kx={{(x-1)}^{2}}$
As per identity rule \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] ,we can expand ${{(x-1)}^{2}}$
$\Rightarrow kx={{x}^{2}}-2x+1$
Now we will rearrange the equation for getting all the $x$ terms at one side,
$\Rightarrow {{x}^{2}}-2x-kx+1=0.........(iii)$
Now, taking x as common and converting the above equation to quadratic form of $a{{x}^{2}}+bx+c=0$
$\Rightarrow {{x}^{2}}-(2+k)x+1=0$
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, $D=0$
Substituting the values from (iv) we get
$\Rightarrow {{[-k-2]}^{2}}-4=0$
$\Rightarrow {{k}^{2}}+4k+4-4=0$
$\Rightarrow {{k}^{2}}+4k=0$
$\Rightarrow k(k+4)=0$
$\Rightarrow k=0;k=-4$
Hence, the final answer is option (e).

Note: There is a possibility of mistake when substituting the values of b in the determinant. ${{b}^{2}}-4ac=0$ if the sign of the coefficient is not put correctly.
Alternatively, we can rewrite, equation (v) as
$\Rightarrow {{(k+2)}^{2}}=4$
Taking square roots on both sides
$k+2=\pm 2$
$\Rightarrow k=0;k=-4$
So, the answer in this case is again the same.