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# help

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The line x + y - 1 = 0 touches the parabola y^2 = kx.  Find the value of k.

May 6, 2020

#1
+652
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$$K = -4$$

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May 6, 2020
#2
+25565
+1

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!

$$\begin{array}{|rcll|} \hline \color{red}k^2+4k &=& 0 \\ k(k+4) &=& 0 \quad | \quad k\ne 0,~ \text{otherwise}~ y^2 = 0 ~\text{or}~ y=0 ~\text{gives no parabola} \\ k+4 &=& 0 \\ \mathbf{k} &=& \mathbf{-4} \\ \hline \end{array}$$

May 7, 2020