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The line y = 12 - 2x is a tangent to two curves. Each curve has an equation of the form y= k + 6 + kx - x^2, where k is a constant. Find the two values of k.

Feb 20, 2019
edited by Guest  Feb 20, 2019

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The line y = 12 - 2x is a tangent to two curves.

Each curve has an equation of the form y= k + 6 + kx - x^2, where k is a constant.

Find the two values of k.

The Point at the tangent:

$$\begin{array}{|rcll|} \hline y = 12-2x &=& k+6+kx-x^2 \\ 12-2x &=& k+6+kx-x^2 \\ x^2-(2+k)x+6-k &=& 0 \\\\ x &=& \dfrac{2+k\pm \sqrt{(2+k)^2-4(6-k) } } {2} \\ x &=& \dfrac{2+k\pm \sqrt{4+4k+k^2-24+4k } } {2} \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{2+k\pm \sqrt{k^2+8k-20 } } {2}} \\ \hline \end{array}$$

The slope at the tangent:

$$\begin{array}{|rcll|} \hline y &=& k+6+kx-x^2 \\ y' &=& k-2x \quad | \quad \text{The slope of the line is } -2 \\ -2 &=& k-2x \\ \mathbf{k} & \mathbf{=} & \mathbf{ 2x-2} \quad & \quad \mathbf{x=\dfrac{2+k\pm \sqrt{k^2+8k-20 } } {2}} \\\\ k & = & 2+k\pm \sqrt{k^2+8k-20 } -2 \\ \sqrt{k^2+8k-20 } &=& 0 \\ k^2+8k-20 &=& 0 \\ k &=& \dfrac{-8\pm \sqrt{64-4\cdot (-20) } } {2} \\ k &=& \dfrac{-8\pm \sqrt{144} } {2} \\ k &=& \dfrac{-8\pm 12} {2} \\\\ k_1 &=& \dfrac{-8+ 12} {2} \\ \mathbf{k_1} & \mathbf{=} & \mathbf{2} \\\\ k_2 &=& \dfrac{-8- 12} {2} \\ \mathbf{k_2} & \mathbf{=} & \mathbf{-10} \\ \hline \end{array}$$

The two values of k are 2 and -10

Feb 20, 2019