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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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).

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  • 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
we get $\int\large\frac{ dy}{(y+3)}=$$2\int\large\frac{dy}{x+4}$
or $\log (y+3)=2\log (x+4)+\log c$
$\log (y+3)=\log c . (x+4)^2$
$y+3=c(x+4)^2$
Step 3:
It is given that the curve passes through the point $-2,1$
Substituting for $x$ and $y$ in the general equation to evaluate for $c$ we get
$1+3=c(-2+4)^2$
$4=4c$
$c=1$
substituting this for $c$
we get $y+3=(x+4)^2$
This is the required equation of the curve
answered Aug 14, 2013 by sreemathi.v
 

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