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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).

18.    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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Slope m of  line joining (x,y) and (-4,-3) is \frac{y+3}{x+4}

According to the question,

\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\ \implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\ \implies \log (y+3) = 2\log (x+4) + \log k \\ \implies (y+3) = k(x+4)^2

Now, Since the curve passes through (-2,1)

x = -2 , y =1

\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1

Putting the value of k, we get

\\ \implies y+3 = (x+4)^2

Posted by

HARSH KANKARIA

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