# Common tangents to circle $x^2+y^2=\frac{1}{2}$ and parabola $y^2=4x$

I'm having trouble with this. What i do is say $\epsilon: y=mx+b$ is the tangent and it meets the circle at $M_1(x_1,y_1)$, i equate the $y$ of the tangent with the circle: $y=\pm \sqrt{1/2-x^2}$ and then the same with the parabola at $M_2(x_2,y_2)$, but i can't reach a result. I've also tried with this version of the tangent to the circle: $xx_1+yy_1=1/2$. By theory we know that a conic section has a tangent at a given point when the discriminant is zero when we equate the two. I'm very confused. I don't know how to solve this. If someone could help i would be very grateful. Thanks in advance.

Edit

Below i add a figure of the graph i made with Matlab

• From $\epsilon$ being a tangent to $y=f(x)$ at $(x_1,y_1)$ for some $f$ you not only have $$y_1=f(x_1),$$ but also $$m=\frac{\mathrm df(x)}{\mathrm dx}\Bigg|_{x=x_1}.$$ Commented Jun 13, 2016 at 13:27
• Please find tangent concept here : archive.org/details/elementsofcoordi00lone Commented Jun 13, 2016 at 13:40
• I've added a newer response on my answer below, which is neater compared to my previous one. Commented Jun 17, 2016 at 16:45

## 4 Answers

It is useful to make a sketch to see what is going on. Added: Nice picture, you can see that there are two common tangent lines, that are symmetrical about the $x$-axis.

Let $(a,b)$ be the point of tangency to the circle. Then the tangent line has equation $ax+by=1/2$. To find the point(s) of intersection of this tangent line with the parabola, we solve $y^2=\frac{2-4by}{a}$ or equivalently $$ay^2+4by-2=0.$$ For tangency to the parabola, the above equation has a double root, so the discriminant $16b^2+8a$ is $0$.

We now have the system of equations $a^2+b^2=1/2$, $2b^2=-a$. If we eliminate $b^2$, we get $2a^2-a-1=0$. This has the roots $a=1$ (irrelevant) and $a=-\frac{1}{2}$.

• I don't understand how you found $y^2=\frac{2-4by}{a}$ Commented Jun 13, 2016 at 15:00
• I skipped an algebra step or two. We have $ax+by=1/2$, so $x=\frac{1/2-by}{a}$, so $4x=\frac{2-4by}{a}$. But $y^2=4x$, and now we have the equation you quoted. Commented Jun 13, 2016 at 15:04
• Thanks. I got the correct result. But i'm trying to understand. So you link the equation $Δ=0$ with the equation of the circle to find what you want. How can you do that? Are these two equations related? I'm asking whether you could explain the mathematical reasoning that you use to connect these two equations together to find what you want. For example i would do that with the tangent line and the circle because they intersect, i.e. they have something in common. Same thing for the tangent line and the parabola. But it seems that you can connect whatever you want? Commented Jun 13, 2016 at 15:31
• If we could find $a$ and $b$ then we would know the equation(s) of the common tangent line(s). So I try to find the relationship between $a$ and $b$ if the line $ax+by=1/2$ is to be tangent to the parabola. This line is tangent to the parabola if it meets the parabola at a "double point." The main work was finding that relationship, which turned out to be $16b^2=-8a$. But we know that $a^2+b^2=1/2$ so now we can find $a$ and $b$. Commented Jun 13, 2016 at 15:37

Using the tangent equations here we have: \begin{align} &\text{Parabola}:& y^2&=4x\\ &\text{ Tangent at }P(p^2, 2p): &y\cdot 2p&= 2(x+p^2) \\ & &\Rightarrow\quad x-py+p^2&=0\\& &\\ \end{align}

For this line to be a tangent to the circle $x^2+y^2=\frac12$, its distance from $(0,0)$ must equal the radius of the circle $\frac 1{\sqrt2}$. \begin{align} \frac {p^2}{\sqrt{1^2+p^2}}&=\frac 1{\sqrt2}\\ 2p^4-p^2-1&=0\\ (2p^2+1)(p^2-1)&=0\\ \because p^2>0\qquad\therefore p^2&=1\\ p&=\pm 1 \end{align}

Hence equation of common tangent is $$\color{red}{y=\pm (x+1)}\qquad\blacksquare$$

The corresponding points of tangency on the parabola are $(1, \pm 2)$.

It can be asily shown that the corresponding points of tangency on the circle are $(-\frac 12, \pm \frac 12)$.

NB - This solution does not require setting the discriminant of the quadratic to zero for tangency. $$\\ \\ \\\\ \\$$

Let $P = (h,k)$ be a point on the circle $x^2 + y^2 = \dfrac 12$. $\left( \text{Then}\; h^2 + k^2 = \dfrac 12 \right)$. The origin, $O = (0,0)$ is the center of the circle. So the equation of the line $\overleftrightarrow{OP}$ is $kx - hy = 0$

A line tangent to the line $kx - hy = 0$ must have an equation of the form $hx + ky = C$ for some number $C$. Since we want the line to pass through the point $(h,k)$, then $C = h^2 + k^2 = \dfrac 12$. So the equation of the line tangent to the circle $x^2 + y^2 = \dfrac 12$ at the point $(h,k)$ is $hx + ky = \dfrac 12$.

This will intersect the parabola $y^2 = 4x$ when

\begin{align} y^2 &= 4x \\ hx + ky &= \dfrac 12\\ \hline \dfrac h4 y^2 + ky &= \dfrac 12\\ hy^2 +4ky-2 &= 0\\ \end{align}

This is a quadratic equation. The line will be tangent to the parabola only if this equation has exactly one solution. That means that the discriminant must be equal to $0$.

\begin{align} B^2 - 4AC &= 0 \\ 16k^2 + 8h &= 0 \\ h = -2k^2 \end{align}

We solve for $h$ and $k$

\begin{align} h^2 + k^2 &= \dfrac 12 \\ h &= -2k^2 \\ \hline 4k^4 + k^2 &= \dfrac 12\\ k^4 + \dfrac 14 k^2 &= \dfrac 18 \\ k^4 + \dfrac 14 k^2 + \dfrac{1}{64}&= \dfrac{9}{64} \\ k^2 + \dfrac 18 &= \pm \dfrac{3}{8} \\ k &= \pm \dfrac 12 \end{align}

This gives us the four candidates $(h,k) = \left( \pm \dfrac 12, \pm \dfrac 12\right)$, wich corresponds to the four possible tangent lines $y = \pm x \pm 1$. All four of these lines will be tangent to the circle and will intersect the parabola in only one point. But only two of them are going to be tangent to the parabola.

A quick check show that the lines $y = -x -1$ and $y = x + 1$ are the tangent lines.

The tangent to the parabola at the point $(t^2,2t)$ is $x-t^2=t(y-2t)$, because the derivative of $f(y)=y^2/4$ computed at $2t$ is $f'(2t)=t$. Thus $$ty-x-t^2=0$$ Such a line must be tangent to the circle, so $$\frac{|t\cdot0-0-t^2|}{\sqrt{1+t^2}}=\frac{1}{\sqrt{2}}$$ that becomes $$2t^4=1+t^2$$ giving $t^2=1$. Thus the common tangents are $$y-x-1=0$$ and $$-y-x-1=0$$

If you don't know calculus, you can still compute the tangent to the parabola $4x=y^2$ at its point $(t^2,2t)$ by equating \begin{cases} x-t^2=m(y-2t) \\ 4x=y^2 \end{cases} so you get $$4my-8mt+4t^2=y^2$$ so $$y^2-4my+8mt-4t^2=0$$ and the discriminant is zero for $$16m^2-4(8mt-4t^2)=0$$ so for $m=t$.

Note: using the standard $y=mx+q$ just complicates the computations. I used the form $x=my+q$ that's as good.