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

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope $\left(m_{1}\right)$ of the line segment joining $(x, y)$ and $(-4,-3)$ is $\frac{y+3}{x+4}$.

We know that the slope of the tangent to the curve is given by the relation,

$\frac{d y}{d x}$

$\therefore$ Slope $\left(m_{2}\right)$ of the tangent $=\frac{d y}{d x}$

According to the given information:

$m_{2}=2 m_{1}$

$\Rightarrow \frac{d y}{d x}=\frac{2(y+3)}{x+4}$

$\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4}$

Integrating both sides, we get:

$\int \frac{d y}{y+3}=2 \int \frac{d x}{x+4}$

$\Rightarrow \log (y+3)=2 \log (x+4)+\log \mathrm{C}$

$\Rightarrow \log (y+3) \log \mathrm{C}(x+4)^{2}$

$\Rightarrow y+3=\mathrm{C}(x+4)^{2}$                ....(1)

This is the general equation of the curve.

It is given that it passes through point (–2, 1).

$\Rightarrow 1+3=C(-2+4)^{2}$

$\Rightarrow 4=4 C$

$\Rightarrow C=1$

Substituting C = 1 in equation (1), we get:

$y+3=(x+4)^{2}$

This is the required equation of the curve.