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Exercise 9.3 · Q18

Q.At any point (x,y)(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)(-4, -3). Find the equation of the curve given that it passes through (−2,1)(-2, 1).

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Model the condition as dydx=2(y+3)x+4\frac{dy}{dx}=\frac{2(y+3)}{x+4}, separate and integrate to y+3=A(x+4)2y+3=A(x+4)^2, then (−2,1)(-2,1) fixes A=1A=1:   y+3=(x+4)2.\;y+3=(x+4)^2.

1. Translate the words into a DE

At a point (x,y)(x,y):

  • slope of the tangent =dydx=\dfrac{dy}{dx};
  • slope of the line joining (x,y)(x,y) to (−4,−3)(-4,-3) =y−(−3)x−(−4)=y+3x+4=\dfrac{y-(-3)}{x-(-4)}=\dfrac{y+3}{x+4}.

'The tangent slope is twice the segment slope' means

dydx=2⋅y+3x+4.\frac{dy}{dx}=2\cdot\frac{y+3}{x+4}.

2. Separate the variables

dyy+3=2 dxx+4.\frac{dy}{y+3}=\frac{2\,dx}{x+4}.

3. Integrate

log⁡∣y+3∣=2log⁡∣x+4∣+C=log⁡(x+4)2+C.\log|y+3|=2\log|x+4|+C=\log(x+4)^2+C.

Exponentiate (absorb the sign into the constant AA):

y+3=A (x+4)2.y+3=A\,(x+4)^2.

4. Apply the point (−2,1)(-2,1) …

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