Exercise 10.3.7

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Consider the parabolas $\left (y-\frac{1}{2}aight)^2 = 2bx$ and $y = \frac{1}{2} x^2$ from (10.9).
\be
\item[(a)] Show that the first parabola has focus $\left(\frac{1}{2}b,\frac{1}{2}aight)$ and directrix $x=-\frac{1}{2} b$.
\item[(b)] Show that the second parabola has focus $\left(0,\frac{1}{2}ight)$ and directrix $y=-\frac{1}{2}$.
\ee
Hence the focus and directrix of the first parabola are defined over any subfield of $\R$ containing $a$ and $b$. For the second, this is true over any subfield of $\R$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
\item[(a)]
Let ${\mathscr P} _1$ the parabola with focus $F\left(\frac{1}{2}b,\frac{1}{2}aight)$ and directrix $D : x=-\frac{1}{2} b$. Let $M(x,y)$ a point of the plane and $H(-\frac{1}{2}b,y)$ the orthogonal projection of $M$ on $D$. Then
\begin{align*}
M(x,y) \in {\mathscr P} _1 &\iff MF^2 = MH^2\
& \iff \left(x-\frac{b}{2}ight)^2 + \left ( y -\frac{a}{2} ight)^2 = \left(x+\frac{b}{2}ight)^2\
&\iff  \left ( y -\frac{a}{2} ight)^2 = 2bx.
\end{align*}
So the first parabola $\left (y-\frac{1}{2}aight)^2 = 2bx$ has focus $\left(\frac{1}{2}b,\frac{1}{2}aight)$ and directrix $x=-\frac{1}{2} b$.

\item[(b)] Let ${\mathscr P} _2$ the parabola with focus $F\left(0,\frac{1}{2}ight)$ and directrix $D : y=-\frac{1}{2} $. Let $M(x,y)$ a point of the plane and $H(x,-\frac{1}{2})$ the orthogonal projection of $M$ on $D$. Then
\begin{align*}
M(x,y) \in {\mathscr P} _2 &\iff MF^2 = MH^2\
& \iff x^2 + \left ( y -\frac{1}{2} ight)^2 = \left(y+\frac{1}{2}ight)^2\
&\iff x^2 = 2y.
\end{align*}
So the second parabola $y = x^2/2$ has focus $\left(0,\frac{1}{2}ight)$ and directrix $y=-\frac{1}{2}$.
\end{proof}

Exercise 10.3.7

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Exercise 10.3.7

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