+26387 The line \(x + y - 1 = 0\) touches the parabola \(y^2 = kx\).
Find the value of \(k\).
\(\begin{array}{|rcll|} \hline \mathbf{x + y - 1} &=& \mathbf{0} \\ x+y &=& 1 \\ \mathbf{x} &=& \mathbf{1-y} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{y^2} &=& \mathbf{kx} \quad | \quad \mathbf{x= 1-y} \\ y^2 &=& k(1-y) \\ y^2 &=& k-ky \\ y^2 +ky-k &=& 0 \\\\ y &=& \dfrac{-k\pm \sqrt{k^2-4(-k)} }{2} \\ y &=& \dfrac{-k\pm \sqrt{\color{red}k^2+4k} }{2} \\ \hline \end{array} \)
Since the tangent we are looking for has exactly one point in common with the parabola,
this quadratic equation can only have one solution.
So its discriminant must be zero!
\(\begin{array}{|rcll|} \hline \color{red}k^2+4k &=& 0 \\ k(k+4) &=& 0 \quad | \quad k\ne 0,~ \text{otherwise}~ y^2 = 0 ~\text{or}~ y=0 ~\text{gives no parabola} \\ k+4 &=& 0 \\ \mathbf{k} &=& \mathbf{-4} \\ \hline \end{array}\)
