Answer
(a) $$y=\frac{3}{4}x-6.25$$
(b) $$A_1(-3, 4)$$
Work Step by Step
(a) Let's consider point $O(0,0)$ and $A(3,-4)$
$m_{OA}=\frac{-4-0}{3-0}=-\frac{4}{3}$
This is slope of a radius and it is perpendicular to the tangent line, which means that (Let the tangent line be $t$):
$m_{OA}\times m_t=-1$
$-\frac{4}{3}\times m_t=-1$
$m_t=\frac{3}{4}$
So, we must find equation to the line that passes through $(3,-4)$ and has slope $\frac{3}{4}$
$y-y_0=m(x-x_0)$
$y-(-4)=\frac{3}{4}(x-3)$
$y+4=\frac{3}{4}x-\frac{9}{4}$
$y=\frac{3}{4}x-6.25$
(b) The tangent line we found in (a) will be parallel to the tangent line at point $A_1$ that is symmetric to $A$ about the origin. The point will be $A_1(-3, 4)$
