Find the common tangent to the parabolas $y=x^{2}-6x+9$ and $y=-x^{2}-2x-1$.

Question

My approach: I differentiated both equations, and took $x_{1}$ to be the $x$-coordinate of the point that satisfies the first equation, and took $x_{2}$ to be the x-coordinate of the point that satisfies the second equation. I then got $x_{1}+x_{2}=2$. I am unable to proceed further. Thanks for any further help!

Answer 1

If you draw the two parabolas, one of the two common tangents will jump out at you.

But there is another one. You can find both by writing all the equations (not only the second one, which is indeed equivalent to $x_1+x_2=2$): $$\begin{align}&\frac{y_2-y_1}{x_2-x_1}=2x_1-6=-2x_2-2\\\iff&x_2=2-x_1\text{ and }-(x_2+1)^2-(x_1-3)^2=(2x_1-6)(x_2-x_1)\\\iff&x_2=2-x_1\text{ and }-2(x_1-3)^2=4(x_1-3)(1-x_1)\\\iff&(x_1,x_2)=(3,-1)\text{ or }(-1,3).\end{align}$$ The two common tangents are therefore the $x$-axis and the line $y=8(1-x).$

Answer 2

Your equations are $$y = (x - 3)^2$$ $$y = -(x + 1)^2$$ So by symmetry, the tangent line has to pass through the midpoint of the vertexes of the parabola, $([3, 0] + [-1, 0])/2 = [1, 0]$. So the tangent line is some $$y = m(x - 1)$$ Solving the above with either quadratic for $x$ gives $$x = \frac { \cdots \pm \sqrt{m^2 + 8m}}{\cdots}$$ Since the tangent line can only intersect the parabolas at a single point, the discriminant must be zero, so $$m^2 + 8m = 0$$ which gives the 2 tangent lines.

Answer 3

Joachimsthal's notation

See this question and also Joachimsthal's notation for a different way to find the tangent of a conic section (parabola, hyperbola, ellipse). We extend this slightly to consider common tangents of two conics. A brief summary of the linked material follows.

Consider the conic with equation $Ax^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H = 0$; writing the left-hand side as $s(x,y)$ this is just $s(x,y)=0$ or $s=0$ for short. Now let us denote

$$s_{ij}=Ax_ix_j + B(x_iy_j + x_jy_i) + Cy_iy_j + F(x_i + x_j) + G(y_i + y_j) + H$$

for two points $(x_i,y_i)$ and $(x_j,y_j)$ which do not necessarily lie on the conic. In particular, if the two points are identical we have

$$s_{ii} = Ax_i^2 + 2Bx_iy_i + Cy_i^2 + 2Fx_i + 2Gy_i + H$$

and the condition that $(x_i,y_i)$ lies on the conic is just $s_{ii} = 0$.

Moreover, the line joining $(x_i,y_i)$ and $(x_j,y_j)$ is tangent to the conic if and only if $s_{ij}^2 = s_{ii}s_{jj}$. This is the key point, whose proof attached at bottom of answer.

In particular, it's common to take $(x_i,y_i) = (x_1, y_1)$ as a point on the conic at which we want to find the tangent. Then the condition that $(x_j,y_j)$ is a point on the tangent to the conic at $(x_1,y_1)$ is just $s_{1j}^2 = s_{11}s_{jj}$ and since $s_{11}=0$ (the point of tangency is on the conic) this becomes $s_{1j}=0$. As $(x_j,y_j)$ represents a general point on the tangent line, it's standard to drop the suffix $j$ and call the point $(x,y)$ instead. Similarly we abbreviate $s_{1j}$ to $s_1$, to indicate the first point in $s_{ij}$ is the given point of tangency $(x_1,y_1)$ and the second point is kept general.

This is the well-known result that if we take the tangent to the conic $s=0$ at a point $(x_1,y_1)$ on the conic, then the equation of the tangent is $s_1 = 0$, i.e.

$$Ax_1x + B(x_1 y + x y_1) + Cy_1y + F(x_1 + x) + G(y_1 + y) + H = 0$$

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Application to this question

Write the parabolas as

\begin{align} s(x,y) &= x^2 - 6x - y + 9 = 0 \\ t(x,y) &= x^2 + 2x + y + 1 = 0 \end{align} We seek a point $P$ on the first parabola (so $s_{PP}=0$) and $Q$ on the second parabola (so $t_{QQ}=0$) such that the line $PQ$ is tangent to both the first parabola (so $s_{PQ}^2 = s_{PP}s_{QQ}$) and the second (so $t_{PQ}^2 = t_{PP}t_{QQ}$). Since $s_{PP}=t_{QQ}=0$ the common tangency condition is just $s_{PQ} = t_{PQ} = 0$. So we have four equations to solve simultaneously to find $P$ and $Q$:

\begin{align} x_P^2 + 6x_P - y_P + 9 &= 0 \tag{1} \\ x_Q^2 + 2x_Q + y_Q + 1 &= 0 \tag{2} \\ x_Px_Q - 3(x_P + x_Q) - \frac{1}{2}(y_P + y_Q) + 9 &= 0 \tag{3} \\ x_Px_Q + (x_P + x_Q) + \frac{1}{2}(y_P + y_Q) + 1 &= 0 \tag{4} \end{align}

For example,

\begin{align} ((3)+(4))/2 \implies x_Px_Q - (x_P+x_Q) + 5 &= 0 \tag{5} \\ (4)-(3) \implies 4(x_P + x_Q) + (y_P + y_Q) &= 8 \tag{6} \end{align}

We use $(1)$ and $(2)$ to put these entirely in terms of $x$ coordinates, \begin{align} y_P + y_Q &= (x_P^2-6x_P+9) + (-x_Q^2 -2x_Q - 1) \\ \implies y_P + y_Q &= x_P^2 - x_Q^2 - 6x_P - 2x_Q + 8 \tag{7} \end{align}

Substituting $(7)$ into $(6)$ gives

\begin{align} x_P^2 - x_Q^2 -2x_P + 2x_Q &= 0 \\ \implies (x_P-x_Q)(x_P+x_Q) - 2(x_P - x_Q) &= 0 \\ \implies (x_P - x_Q)(x_P + x_Q -2) &= 0 \end{align}

So either $x_P = x_Q$ or $x_P + x_Q = 2$. But putting the former into $(5)$ gives $x_P^2 -2x_P + 5=0$ with no real solutions. The latter gives $x_P(2-x_P) + 3=0$ with solutions $x_P=-1$ or $x_P=3$.

We deduce the two tangent lines $PQ$ are between $P=(-1, 16)$, $Q=(3, -16)$ and $P=(3, 0)$, $Q=(-1, 0)$.

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Proof of tangency condition $s_{ij}^2 = s_{ii}s_{jj}$

Write the vector equation of the line in parametric form, $(x,y)=t(x_i,y_i) + (1-t)(x_j,y_j)$. When the parameter $t=0$ we are at $(x_j,y_j)$ and when $t=1$ we are at $(x_i,y_i)$.

To find the line's intersection points with the conic, substitute $(x,y)$ into our expression for $s(x,y)$ so we express $s$ in terms of the parameter $t$. The equation of the conic will be satisfied when $s(t)=0$; this turns out (for a fuller algebraic proof, see this answer) to yield the quadratic equation

$$(s_{ii}-2s_{ij}+s_{jj})t^2 + 2(s_{ij}-s_{jj})t + s_{jj} = 0$$

with discriminant $$\Delta = 4(s_{ij}-s_{jj})^2 - 4(s_{ii}-2s_{ij}+s_{jj})s_{jj}$$ which simplifies to $4(s_{ij}^2 - s_{ii}s_{jj})$. For a tangent, we want the line to touch the conic only once, i.e. one solution to the quadratic, so the discriminant must be zero. This happens when $s_{ij}^2 = s_{ii}s_{jj}$.

Answer 4

Let $y=ax+b$ be such a tangent. This line will intersect the parabolae at points respectively given by:

$x^2-6x+9=ax+b\implies x^2-(a+6)x+(-b+9)=0$

$-x^2-2x-9=ax+b\implies x^2-)+(a+2)x+(b+1)=0$

For tangency at either parabola the corresponding quadratic equation must have a double root and so its discriminant must be zero. For a common tangent we must therefore have both:

$(a+6)^2-4(-b+9)=a^2+12a+4b=0$

$(a+2)^2-4(b+1)=a^2+4a-4b=0$

Taking the sum and difference between these equations and dividing out common factors gives respectively

$a^2+8a=0\implies a\in\{-8,0\}$

$a+b=0\implies b=-a$

from which the solution for the two tangent lines follows.

Answer 5

A common tangent has the equation $y = l(x)$, with $(x-3)^2 - l(x) = l^2_1(x)$, and $l(x) - (-(x+1)^2) = l_2^2(x)$. Adding up we get

$$(x-3)^2 + (x+1)^2 = l_1(x)^2 + l_2(x)^2$$

with 2 solutions for $(l^2_1(x), l^2_2(x))$ being

$$(l^2_1(x), l^2_2(x)) = ((x-3)^2, (x+1)^2) \\ (l^2_1(x), l^2_2(x)) = ((x+1)^2, (x-3)^2 )$$

The two common tangent are given by $y = 0$, and $y = l(x) = (x-3)^2 - (x+1)^2 = -8 x + 8$

Answer 6

The first parabola is

$y = x^2 - 6 x + 9$

The second parabola is

$ y = - x^2 - 2 x - 1 $

Place a point $P$ on the first parabola, then

$ P = (x, x^2 - 6 x + 9 )$

The tangent vector to $P$ is $ d = (1, 2 x - 6) $

Therefore, the tangent line has a parametric equation

$ Q = P + t d = (x, x^2 - 6 x + 9) + t (1, 2x - 6) = (x + t , x^2 - 6 x + 9 + t (2 x - 6) ) $

Intersect this tangent line with the second parabola then

This gives us,

$ x^2 - 6 x + 9 + t (2 x - 6) = - (x + t)^2 - 2 (x + t) - 1 $

Expand,

$ x^2 - 6 x + 9 + 2 t x - 6 t = - x^2 - 2 x t - t^2 - 2 x - 2 t - 1 $

Simplify,

$2 x^2 - 4 x + 10 + 4 t x - 4 t + t^2 = 0 $

This is a quadratic equation in $t$. If it has a single solution then the discriminant must be zero, i.e.

$ (4 x - 4)^2 - 4 (2 x^2 - 4 x + 10) = 0 $

Expanding,

$ 16 x^2 - 32 x + 16 - 8 x^2 + 16 x - 40 = 0 $

Simplifying,

$ 8 x^2 - 16 x - 24 = 0 $

Simplify,

$ x^2 - 2 x - 3 = 0 $

Factor,

$(x - 3)(x + 1) = 0 $

This gives the two possible tangents:

$P_1 = ( 3, 0 ) , \ d_1 = (1, 0 ) $

$ P_2 = (-1, 16 ) , \ d_2 = (1, -8 ) $

Hence, the two tangents are

$ p_i(t) = (x, y) = P_i + t \ d_i $

Let $u_i \perp d_i$ then $u_1 = (0, 1), \ u_2 = (8, 1) $

then we get the Cartesian equation by dotting $p_i$ with u_i$

$ u_i \cdot (x, y) = u_i \cdot P_i $

For the first tangent, we get

$ y = 0 $

And for the second tangent , we get

$ 8 x + y = 8 $