Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
A reason why you are down-voting this question would be appreciated. –  infinitesimal simplicio 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

up vote 4 down vote accepted

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

share|improve this answer

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 $k$.

share|improve this answer

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 $0$

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

share|improve this answer

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}. $

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.