Exercise 10.3.4

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}}

Show that the tangent line at a point $(x_1,y_1)$ on the first parabola in (10.9) has slope given by
$$m=\frac{b}{y_1-\frac{1}{2}a}.$$

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}The equation of the tangent $T$ at the point $M_1(x_1,y_1)$ to the parabola whose equation is
$$f(x,y) = 2bx - \left(y-\frac{1}{2}aight)^2 =0,$$
is
\begin{align*}
0 &=\frac{\partial f}{\partial x}(x_1,y_1) (x-x_1) + \frac{\partial f}{\partial y}(x_1,y_1) (y-y_1),\
&= 2b (x-x_1) -  2\left(y_1-\frac{1}{2}aight) (y-y_1),
\end{align*}
so the equation of $T$ is
$$T : bx - \left(y_1-\frac{1}{2}aight)y -bx_1+\left(y_1-\frac{1}{2}aight)y_1 = 0.$$
The slope of $T$ is 
$$m=\frac{b}{y_1-\frac{1}{2}a}.$$
\end{proof}

Exercise 10.3.4

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Exercise 10.3.4

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