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 $(-2,1)$.

Answer 1

As we know that,

Slope of tangent to the curve is $\frac{dy}{dx}$

Slope of the line segement joining the two points$(x,y)\text{and}(-4,-3)$ is

$\frac{y_2-y_1}{x_2-x_1}=\frac{-3-y}{-4-x}=\frac{y+3}{x+4}$

Given : Slope of tangent is twice of line segment

$\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

$\frac{dy}{y+3}=\frac{2dx}{x+4}$

Integrating both sides, we get

$\int \frac{dy}{y+3}=2\int \frac{dx}{x+4}$

$\log |y+3|=2\log |x+4|+\log c$

$\log (y+3)=\log (x+4{)}^2+\log c$

$\log (y+3)-\log (x+4{)}^2=\log c$

$\log \frac{(y+3)}{(x+4{)}^2}=\log c$

$c=\frac{y+3}{(x+4{)}^2}$ ...$\left(i\right)$

Given : Curve passes through $(2,-1)$

Substituting $x=-2,y=1$ in $\left(i\right)$,

$c=\frac{1+3}{(-2+4{)}^2}$

$c=\frac{4}{(2{)}^2}=\frac{4}{4}=1$

Substituting the value of $c$ in $\left(i\right)$,

$\frac{y+3}{(x+4{)}^2}=1$

$y+3=(x+4{)}^2=1$

Hence, required equation of the curve is

$y+3=(x+4{)}^2$