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 $( - 2,1)$.
Differential Equations — Class 12 Maths Solution
Step-by-step Solution
:
We are given that $\cfrac{{dy}}{{dx}} = 2\left( {\cfrac{{y + 3}}{{x + 4}}} \right) \Rightarrow \cfrac{{dy}}{{y + 3}} = 2\cfrac{{dx}}{{x + 4}}$
… (1)
Integrating (1) both sides,
we get
$\int {\cfrac{{dy}}{{y + 3}}} = 2\int {\cfrac{{dx}}{{x + 4}}} \Rightarrow \log (y + 3) = 2\log (x + 4) + C$
As point $( - 2,1)$ lies on it
So, $\log 4 = 2\log 2 + C \Rightarrow 2\log 2 = 2\log 2 + C \Rightarrow C = 0$
$\therefore$ Equation of the curve is
$\log (y + 3) = 2\log (x + 4)$
$\Rightarrow y + 3 = {(x + 4)^2}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.