Question

Помогите пожалуйста ! Помогите пожалуйста

Answer 1

$y=ax^2+bx+c\ \ ,\ \ A\Big(-\dfrac{1}{9}\, ;\, 0\Big)\ \ ,\ \ B\Big(\, \dfrac{1}{4}\, ;\, 0\Big)\ \ ,\ \ C(\, 0\, ;\, c)\ ,\ c<0\ \ ,\ \ AC\perp\ BC\\\\\\A\Big(-\dfrac{1}{9}\, ;\, 0\Big):\ \ \ \dfrac{1}{81}\cdot a-\dfrac{1}{9}\cdot b+c=0\ \ ,\ \ a-9b+81c=0\\\\B\Big(\, \dfrac{1}{4}\, ;\, 0\Big):\ \ \ \ \ \ \dfrac{1}{16}\cdot a+\dfrac{1}{4}\cdot b+c=0\ \ ,\ \ a+4b+16c=0$

$\left\{\begin{array}{ccc}a-9b+81c=0\\a+4b+16c=0\end{array}\right\ \ominus \ \left\{\begin{array}{l}13b-65c=0\\a+4b+16c=0\end{array}\right\ \ \left\{\begin{array}{l}\ b=\dfrac{65}{13}\, c\\a+\dfrac{4\cdot 65}{13}\, c+16c=0\end{array}\right\\\\\\\left\{\begin{array}{l}b=5\, c\\a=-20c-16c\end{array}\right\ \ \left\{\begin{array}{l}b=5c\\a=-36c\end{array}\right$

$AC\perp\ BC\ \ \Rightarrow \ \ k_{AC}\cdot k_{BC}=-1\ \ ,\ \ k=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\k_{AC}=\dfrac{c-0}{0+\frac{1}{9}}=9c\ \ ,\ \ k_{BC}=\dfrac{c-0}{0-\frac{1}{4}}=4c\\\\\\k_{AC}\cdot k_{BC}=-9c\cdot 4c=-1\ \ \ \Rightarrow \ \ \ 36c^2=1\ \ ,\ \ c^2=\dfrac{1}{36}\ \ ,\ \ c=\pm \dfrac{1}{6}\\\\c<0\ \ \Rightarrow \ \ c=-\dfrac{1}{6}\\\\a=-36\cdot c=-36\cdot \Big(-\dfrac{1}{6}\Big)=6\\\\\\\boxed{\ a=6\ }$

$P.S.\ \ y=6x^2+bx-\dfrac{1}{6}\\\\b=5\, c=-\dfrac{5}{6}\\\\\\y=6x^2-\dfrac{5}{6}\cdot x-\dfrac{1}{6}$