Phương trình $(d)$:
$y=\dfrac{-1}{(2a+3)^2}(x-a)+\dfrac{a+2}{2a+3}$
Thay $x=0$ vào phương trình $(d)$, ta có:
$y=\dfrac{-1}{(2a+3)^2}.(-a)+\dfrac{a+2}{2a+3}$
$=\dfrac{a}{(2a+3)^2}+\dfrac{(a+2)(2a+3)}{(2a+3)^2}$
$=\dfrac{a+2a^2+7a+6}{(2a+3)^2}$
$=\dfrac{2a^2+8a+6}{(2a+3)^2}$
Vậy $d∩Oy=B\Bigg(0;\dfrac{2a^2+8a+6}{(2a+3)^2}\Bigg)$
Thay $y=0$ vào phương trình $(d)$, ta có:
$0=\dfrac{-1}{(2a+3)^2}(x-a)+\dfrac{a+2}{2a+3}$
$⇔ \dfrac{a-x}{(2a+3)^2}+\dfrac{(a+2)(2a+3)}{(2a+3)^2}=0$
$⇔ \dfrac{a-x+2a^2+7a+6}{(2a+3)^2}=0$
$⇔ 2a^2+8a+6-x=0$
$⇔ x=2a^2+8a+6$
Vậy $d∩Ox=A(2a^2+8a+6;0)$
