At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point `("-"4,"-"3)` . Find the equation of the curve given that it passes through `(-2," "1)` .

Explanation

Slope of tangent to the curve `=\frac{d y}{d x}`

Slope of the line segment joining `(x, y)` and `(-4,-3)` is
`\frac{y+3}{x+4}`

Given, at point `(x, y)`.Slope of tangent is twice of line segment

$$
\begin{aligned}
&\frac{d y}{d x}=2\left(\frac{y+3}{x+4}\right) \\
&\frac{d y}{y+3}=\frac{2 d x}{x+4}
...