Browse Questions

# 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).

Toolbox:
• Equation of a line passing througn a given slope and point is $y-y_1 = m(x-x_1)$
Step 1:
It is given that $xy$ is the point of contact of the curve and its tangent.
Hence the slope=$\large\frac{ y+3}{x+4}$.Let this be $m_1$
but we know that slope of the tangent to the curve is $\large\frac{dy}{dx}$.Let this be $m_2$
According to the given information, $m_2= 2m_1$
$\large\frac{dy}{dx}=\frac{ 2(y+3)}{x+4}$
Seperating the variables
we get $\large\frac{dy}{(y+3)} =\frac{ 2dx}{(x+4)}$
Step 2:
Integrating on both sides