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

Question

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

Answer 1

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 $a=-\frac{1}{2}$.

Answer 2

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. $$\\ \\ \\\\ \\ $$

Answer 3

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.

Answer 4

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.