Questão 9 - 1ª Fase - ITA 2026

(i) As coordenadas do vértice $\mathrm{V}$ da parábola $\mathrm{P}: \mathrm{y}={\mathrm{x}}^2+\mathrm{x}+6$ são dadas por:

${\mathrm{x}}_{\mathrm{V}}=\frac{-\mathrm{b}}{2\mathrm{a}}=\frac{-1}{2}$ e ${\mathrm{y}}_{\mathrm{V}}={\left(-\frac{1}{2}\right)}^2+\left(\frac{-1}{2}\right)+6=\frac{23}{4}\to \mathrm{V}=\left(-\frac{1}{2},\frac{23}{4}\right)$

(ii) Sejam ${\mathrm{m}}_{\mathrm{r}}$ e ${\mathrm{m}}_{\mathrm{t}}$ os coeficientes angulares das retas $\mathrm{r}$ e $\mathrm{t}$. Assim:

$\mathrm{r}: 3\mathrm{x}+4\mathrm{y}=-15\leftrightarrow \mathrm{y}=-\frac{3}{4}\mathrm{x}-\frac{15}{4}\to {\mathrm{m}}_{\mathrm{r}}=-\frac{3}{4}$

Como $\mathrm{r}$ e $\mathrm{t}$ formam ângulo agudo de 45º, tem-se:

$\left|\frac{{\mathrm{m}}_{\mathrm{t}}-{\mathrm{m}}_{\mathrm{r}}}{1+{\mathrm{m}}_{\mathrm{r}}·{\mathrm{m}}_{\mathrm{t}}}\right|=\mathrm{tg}45º\leftrightarrow \left|\frac{{\mathrm{m}}_{\mathrm{t}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{3}{4}}·{\mathrm{m}}_{\mathrm{t}}}\right|=1\leftrightarrow \left\{\begin{array}{l}{\mathrm{m}}_{\mathrm{t}}+\frac{3}{4}=1-\frac{3}{4}·{\mathrm{m}}_{\mathrm{t}}\\ {\mathrm{m}}_{\mathrm{t}}+\frac{3}{4}=-1+\frac{3}{4}·{\mathrm{m}}_{\mathrm{t}}\end{array}\right.\leftrightarrow \left\{\begin{array}{l}{\mathrm{m}}_{\mathrm{t}}=\frac{1}{7}\\ {\mathrm{m}}_{\mathrm{t}}=-7\end{array}\right.$

O enunciado pede ${\mathrm{m}}_{\mathrm{t}}$ negativo. Logo:

${\mathrm{m}}_{\mathrm{t}}=-7 \mathrm{e} \mathrm{t}: 7\mathrm{x}+\mathrm{y}+\mathrm{k}=0$

(iii) O coeficiente angular da reta tangente à parábola no ponto A é dado por:

${\mathrm{m}}_{\mathrm{t}}=\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{x}}_{\mathrm{A}}+1\to 2{\mathrm{x}}_{\mathrm{A}}+1=-7\to {\mathrm{x}}_{\mathrm{A}}=-4$

${\mathrm{y}}_{\mathrm{A}}={\left(-4\right)}^2+\left(-4\right)+6=18\to \mathrm{A}=\left(-4,18\right)$

(iv) O ponto $\mathrm{A}=\left(-4,18\right)$ pertence à reta $\mathrm{t}: 7\mathrm{x}+\mathrm{y}+\mathrm{k}=0:$

$7·\left(-4\right)+18+\mathrm{k}=0\leftrightarrow \mathrm{k}=10\to \mathrm{t}: 7\mathrm{x}+\mathrm{y}+10=0$

Determinação do ponto B, que é a interseção de r e t:

$\left\{\begin{array}{l}3\mathrm{x}+4\mathrm{y}=-15\\ 7\mathrm{x}+y=-10\end{array}\right.\leftrightarrow \left\{\begin{array}{l}\mathrm{x}=-1\\ \mathrm{y}=-3\end{array}\right.\to \mathrm{B}=\left(-1,-3\right)$

(v) Cálculo da área do triângulo ABV:

$\mathrm{D}=\left|\begin{array}{cccc}-4& -1& -\frac{1}{2}& -4\\ 18& -3& \frac{23}{4}& 18\end{array}\right|=12-\frac{23}{4}-9+18-\frac{3}{2}+23=\frac{147}{4}$

$\left[\mathrm{ABV}\right]=\frac{\left|\mathrm{D}\right|}{2}=\boxed{\frac{147}{8}}$