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 the contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

Answer 1

It is given that (x,y) is the point of cinatact of the curve and its tangent.
The slope of the line segment joining the points $(x_2,y_2)\to (x,y)$ and $(x_1,y_1)\to (-4,-3)$
$=\frac{y-(-3)}{x-(-4)}=\frac{y+3}{x+4}$ $\therefore Slopeoftangent=\frac{y_2-y_1}{x_2-x_1}$
According to the question (slope of tangent is twice the slope of the line), we must have, $\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$
Now, separating the variables, we get $\frac{dy}{y+3}=\left(\frac{2}{x+4}\right)$
On integarting both sides, we get $\int \frac{dy}{y+3}=\int \left(\frac{2}{x+4}\right)\Rightarrow log|y+3|=2log|x+4|+log|C|\Rightarrow log|y+3|=log|x+4{|}^2+log|C|\Rightarrow log\frac{|y+3|}{|x+4{|}^2}=log|C|\Rightarrow \frac{|y+3|}{|x+4{|}^2}=C$
The curve passes through the point (-2,1), therefore
$\Rightarrow \frac{|1+3|}{|-2+4{|}^2}=C\Rightarrow C=1$
Substituting C=1 in Eq. (ii), we get
$\frac{|y+3|}{|x+4{|}^2}=1\Rightarrow y+3=(x+4{)}^2$
which is the required equation of curve.