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

According to question, we have
$\frac{dy}{dx}=2\frac{y+3}{x+4}\frac{dy}{(y+3)}=2\frac{dx}{(x+4)}ln(y+3)=2ln(x+4)+lncln(y+3)=ln(x+4{)}^2+lnc(y+3)=c.(x+4{)}^2$
and it passes through $(-2,1)$
$(1+3)=c(-2+4{)}^{2}4=c4c=1$
equation of curve is $(y+3)=(x+4{)}^2$