# 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

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 $(-2,1)$.

Solution:

It is given that (xy) 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.