Tangent and Normal Questions in English

(C) The equation of the curve is $xy = 4$,which can be written as $y = \frac{4}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.
The equation of the line is $ax + by + c = 0$,which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $m = -\frac{a}{b}$.
Since the line is a tangent to the curve,the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:
$-\frac{4}{x^2} = -\frac{a}{b} \implies \frac{a}{b} = \frac{4}{x^2}$.
Since $x^2 > 0$ for all $x \neq 0$,it follows that $\frac{a}{b} > 0$.
This implies that $a$ and $b$ must have the same sign.
Looking at the options,the condition $a < 0$ and $b < 0$ satisfies $\frac{a}{b} > 0$.

(C) To find the angle between the curves $y^2=x$ and $x^2=y$ at the origin $(0,0)$:
$1$. For the curve $y^2=x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2y}$. At the origin $(0,0)$,the slope is undefined,which means the tangent is the $y$-axis $(x=0)$.
$2$. For the curve $x^2=y$,differentiating with respect to $x$ gives $2x = \frac{dy}{dx}$. At the origin $(0,0)$,the slope is $\frac{dy}{dx} = 0$,which means the tangent is the $x$-axis $(y=0)$.
$3$. The angle between the $x$-axis and the $y$-axis is $\frac{\pi}{2}$ radians. Therefore,the angle between the two curves at the origin is $\frac{\pi}{2}$.