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.