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

The slope of the line having points (x, y) and (−4, −3) is given by $\frac{y+3}{x+4}$.
According to the question,

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

$$\begin{gathered} \Rightarrow \frac{1}{y+3}dy=\frac{2}{x+4}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y+3}dy=2\int \frac{1}{x+4}dx \\ \Rightarrow \log \left|y+3\right|=2\log \left|x+4\right|+\log C \\ \Rightarrow \log \left|y+3\right|=\log \left|C{\left(x+4\right)}^2\right| \\ \Rightarrow y+3=C{\left(x+4\right)}^2 \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(-2,1\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ \therefore 1+3=C{\left(-2+4\right)}^2 \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get} \\ y+3={\left(x+4\right)}^2 \end{gathered}$$