Question

At any point $(x,y)$ of a curve,the slope of a 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

It is given that $(x,y)$is the point of contact of the curve and its tangent.
The slope $(m_1)$ 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,$\frac{dy}{dx}$
∴Slope $(m_2)$ of the tangent$=\frac{dy}{dx}$
According to the given information:
$m_2=2m_1$
$⇒\frac{dy}{dx}=\frac{2(y+3)}{x+4}$
$⇒\frac{dy}{y+3}=2\frac{dx}{x+4}$
Integrating both sides,we get:
$∫\frac{dy}{y+3}=2∫\frac{dx}{x+4}$
$⇒log(y+3)=2log(x+4)+logC$
$⇒log(y+3)logC(x+4)^2...(1)$
This is the general equation of the curve.
It is given that it passes through point(-2,1).
$⇒1+3=C(-2+4)^2$
$⇒4=4C$
$⇒C=1$
Substituting C=1 in equation(1),we get:
$y+3=(x+4)^2$
This is the required equation of the curve.