Question

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 $(-2,1)$

Answer 1

Slope of tangent to the curve$=\frac{dy}{dx}$

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

$\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

$\frac{dy}{y+3}=\frac{2dx}{x+4}$

Integrating both sides, we get

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

$\log (y+3)=2\log (x+4)+\log C$

$\log (y+3)-\log {(x+4)}^2=\log C$

$\log \frac{y+3}{{(x+4)}^2}=\log C$

$\frac{y+3}{{(x+4)}^2}=C$ ....$\left(1\right)$

The curve passes through $(-2,1)$

Put $x=-2$ and $y=1$ in $\left(1\right)$

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

Put value $C=1$ in eqn$\left(1\right)$ we get

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

Hence, the equation of the curve is $y+3={(x+4)}^2$