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Using the curve $y^2+2x=13$, find the value of $k$ for which the line $2y+x=k$ is a tangent to the curve.

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A reason why you are down-voting this question would be appreciated. –  skullpatrol Dec 20 '12 at 13:40
    
Hint (edit): Derive both equations and equate the slopes, then work out the corresponding point of intersection. Then find k. –  Sp3000 Dec 20 '12 at 13:41

4 Answers 4

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Differentiate $y^2+2x=13$ to get $2y\frac{dy}{dx}+2=0$, so $\frac{dy}{dx}=-\frac{1}{y}$. From $2y+x=k$ we have $y=\frac{1}{2}(k-x)$, so that the gradient is $-\frac{1}{2}$. Hence $y=2$ and $x=\frac{9}{2}$ on the curve when the line is a tangent. Therefore $k=\frac{17}{2}$.

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Using the curve $y^2+2x=13$, find the value of $k$ so the line $2y+x=k$ is tangent to the curve.

I trust you're fully capable of carrying out the computations, so I won't insult you by doing that for you. I'll outline the procedure you can follow so you can generalize to other problems of this sort.

Key words: point-of-intersection, derivative & slope: tangent.

(1) Differentiate $y^2+2x=13$ and solve for $\frac {dy}{dx}$.

(2) From the equation $2y + x = k \iff y = \frac{1}{2}(k - x) = -\frac{1}{2}x + \frac12 k$, determine the slope $m$ of the line.

(3) Substitute slope from (2) into your result from (1) to determine the $y$-coordinate at the point where the line is tangent to the curve (solve for "$y$" by equating "slopes": put $\frac{dy}{dx} = m$).

(4) Use the equation of the curve $(y^2+2x=13)$ and the value of $y$ obtained in (3) to solve for the $x$-coordinate of that point of intersection. Using the values of $x, y$ at the point of intersection: $(x, y)$, substitute those values into the equation $2y + x = k$ to solve for $k$.

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Let us find the intersection of $x+2y=k--->(1)$ and the curve $y^2+2x=13--->(2)$

Replacing $x$ with $k-2y,$ we get $y^2+2(k-2y)=13, y^2-4y+2k-13=0$

This is a quadratic equation in $y$

As $(1)$ a tangent of $(2),$ the point of intersections of will coincide. So, the discriminant $4^2-4(2k-13)=68-8k$ must be equal to $0$

$\implies k=\frac{17}2$

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As $ x = \dfrac{1}{2}(13-y^{2}) $ is a parable and $ x= k-2y $ is a line, an geometric answer is to determine $ k $ such that $ \dfrac{1}{2}(13-y^{2}) =k-2y $ has a unique solution. This implies that $ y^{2}-4y +(2k-13)=0 $ must have only one solution. And this implies that \begin{equation} \Delta = 4^2 - 4(2k-13) = 0. \end{equation} Then $ k = \dfrac{17}{2}. $

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