Exercise 10.3.3

Proof.

(a)

Let $P$ be the parabola with focus $P_1$ and directrix $l_1$.

With the chosen coordinates system, let $Q(x,y)$ a point in the plane, and $Q_1$ the orthogonal projection of $Q$ on the directrix. Knowing $P_1(0,a)$ and $Q_1(x,-a)$, we obtain

$$\begin{array}{llll}\hfill Q\in P& ⟺Q{P}_1^2=Q{Q}_1^2& \hfill & \\ \hfill & ⟺x^2+{(y-a)}^2={(y+a)}^2& \hfill & \\ \hfill & ⟺x^2+y^2-2\mathit{ay}+a^2=y^2+2\mathit{ay}+a^2& \hfill & \\ \hfill & ⟺x^2=4\mathit{ay}& \hfill & \end{array}$$

So the equation of $P$ is

$$P:x^2=4\mathit{ay}.$$

(b)

Let $Q(x_0,y_0)$ be a point on the parabola, so $x_0^2=4a{y}_0$.

Write $F(x,y)=x^2-4\mathit{ay}$. The equation of the tangent $T$ to $P$ at the point $Q$ is given by

$$\begin{array}{llll}\hfill 0& =\frac{\mathit{\partial F}}{\mathit{\partial x}}(x_0,y_0)(x-x_0)+\frac{\mathit{\partial F}}{\mathit{\partial y}}(x_0,y_0)(y-y_0),& \hfill & \\ \hfill 0& =2x_0(x-x_0)-4a(y-y_0),& \hfill & \\ \hfill & 2x_{0}x-4\mathit{ay}=2x_0^2-4a{y}_0=4a{y}_0.& \hfill & \end{array}$$

So the equation of $T$ is

$$T:x_{0}x-2\mathit{ay}=2a{y}_0.$$

Let $R(x_R,y_R)$ the intersection point of $T$ with the $y$-axis.

Then $x_R=0$, so $-2a{y}_R=2a{y}_0,y_R=-y_0$,

$$R(0,-y_0).$$

(c)

Write $\mathit{MN}=||\stackrel{\to }{\mathit{MN}}||$ the length of the segment $\overline{\mathit{MN}}$.

$$P_1(0,a),R(0,-y_0),Q_1(x_0,-a),Q(x_0,y_0),(a>0,y_0>0),$$

thus $\stackrel{\to }{P_{1}R}=(0,-a-y_0)=\stackrel{\to }{Q{Q}_1}$, hence $P_{1}R=Q{Q}_1$. By definition of the parabola, $Q{P}_1=Q{Q}_1$, so

$Q{P}_1=y_0+a=\sqrt{x_0^2+{(y_0-a)}^2}=R{Q}_1$, therefore

$$Q{P}_1=Q{Q}_1=P_{1}R=R{Q}_1:$$

$(Q,P_1,R,Q_1)$ is a rhombus, whose length of side is $c=a+y_0$. Hence the diagonals $(P_1{Q}_1)$ and $T=(\mathit{QR})$ are perpendicular. We give a direct proof: as $\stackrel{\to }{P_{1}R}=\stackrel{\to }{Q{Q}_1}$,

$$\begin{array}{llll}\hfill \stackrel{\to }{\mathit{QR}}.\stackrel{\to }{P_1{Q}_1}& =(\stackrel{\to }{Q{P}_1}+\stackrel{\to }{P_{1}R}).(\stackrel{\to }{P_{1}Q}+\stackrel{\to }{Q{Q}_1})& \hfill & \\ \hfill & =\stackrel{\to }{Q{P}_1}.\stackrel{\to }{P_{1}Q}+\stackrel{\to }{Q{P}_1}.\stackrel{\to }{Q{Q}_1}+\stackrel{\to }{P_{1}R}.\stackrel{\to }{P_{1}Q}+\stackrel{\to }{P_{1}R}.\stackrel{\to }{Q{Q}_1}& \hfill & \\ \hfill & =-c^2+\stackrel{\to }{Q{P}_1}(\stackrel{\to }{Q{Q}_1}-\stackrel{\to }{P_{1}R})+c^2& \hfill & \\ \hfill & =0& \hfill & \end{array}$$

As in any parallelogram, the intersection of the diagonals is the middle point of $(P_1,Q_1)$ and $(Q,R)$, so $Q_1$ is the reflection of $P_1$ about the tangent line to the parabola at $Q$.

(d)

Conversely, let $Q_1(x_0,-a)$ be any point on the directrix $l_1$, and $l$ the perpendicular bisector of $(P_1,Q_1)$. We must prove that $l$ is tangent to the parabola $P$. Let $M(x,y)$ a point of the plane. With $P_1(0,a)$, and $Q_1(x_0,-a)$, $$\begin{array}{llll}\hfill M\in l& ⟺M{P}_1^2=M{Q}_1^2& \hfill & \\ \hfill & ⟺x^2+{(y-a)}^2={(x-x_0)}^2+{(y+a)}^2& \hfill & \\ \hfill & ⟺x^2+y^2-2\mathit{ay}+a^2=x^2-2x_{0}x+x_0^2+y^2+2\mathit{ay}+a^2& \hfill & \\ \hfill & ⟺x_{0}x-2\mathit{ay}=\frac{x_0^2}{2}& \hfill & \end{array}$$

If we write $y_0=\frac{x_0^2}{4a}$, then the point $M_0(x_0,y_0)$ lies on the parabola $P$. Since $x_0^2-2a{y}_0=\frac{x_0^2}{2}$, $M_0$ lies also on $l$. By part (b), the equation of the tangent $T$ at point $M_0$ is $x_{0}x-2\mathit{ay}=2a{y}_0=x_0^2∕2$, hence $T=l$, so $l$ is tangent to the parabola $P$.

Conclusion: the reflexion of $P_1$ about $l$ lies on the line $l_1$ if and only if $l$ is tangent to the parabola with focus $P_1$ and directrix $l_1$.