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I have this question.

Find the equation of the tangent to the line $y^2=x$ at the point $(16,-4)$. I have tried to use both methods to work it out.

1) Substitute $y=mx+c$ into $y^2=x$ and find a quadratic in terms of $x$ then set the discriminant to zero to try and find $m$ but I ended up with an extra $c$ term and got stuck.

2) Use derivatives to find the value of $m$, I did the calculations and also got the wrong answer.

Also could anyone help me extend these ideas to conics generally.

I'm getting extremely frustrated because it seems like it should be easy but I've spent 2 hours working on it and honestly just feel like quitting mathematics all together obviously I'm not cut out for it. $:($

Cheers.

Sean B
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4 Answers4

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You can use the general equation of a line through the point(16, -4). It will intersect the parabola in $(16,-4)$ and another point. The tangent will correspond to the case when this other point is $(16,-4)$ again, which will give a condition on $m$ for this to happen.

A line with slope $m$ passing through $(16,-4)$ has equation $\,\,y+4=m(x-16)$. To find the intersection points of this line with the parabola, you have to solve \begin{align*}\bigl(m(x-16)-4\bigr)^2=x&\iff m^2(x-16)^2-8m(x-16)+16-x=0\\ & (x-16)\bigl(m^2(x-16)-(8m+1)\bigr)=0 \end{align*} One root is $x=16$; the other one is also $x=16$ if and only if $\,\,8m+1=0$, i.e. $\,\,m=-\frac18$, whence the equation of the tangent: $$x+8y+16=0.$$

Bernard
  • 175,478
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Use calculus:

Differentiating gives $\frac{dy}{dx} = \frac1{2y}$. Thus the slope of the line is $\frac{1}{2y}$, which, at (16,-4) is equal to $-\frac{1}{8}$. Using the point-slope form of a line gives us:

$y+4=-\frac18(x-16)$, where we have plugged in the point in question.

Gabriel
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You write $y=m(x-16)-4$ (this gives parametrisation of all lines through $(-4,16)$).

Replacing in $x-y^2$ you want get a polynomial in $x$ which should divide $(x-16)^2$.

$x-y^2=(x-16)(-m^2(x-16)+8m+1)$, so $m=-\frac{1}{8}$, and the tangent line has equation

$$y=-\frac{1}{8}(x-16)-4.$$

Jérémy Blanc
  • 3,939
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You should first find dy/dx in terms of y and x.

$y^2 = x \implies 2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$

Now, the slope of the tangent is $\frac{dy}{dx}$ evaluated at $y=-4$ which is $\frac{-1}{8}$, I think you can do the rest.