54. 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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Vishal Baghel | Contributor-Level 10

8 months ago

The slope of tangent is $\frac{d y}{d x}$ and slope of line joining line (-4,-3) and point say P(x,y)

$\frac{y - (- 3)}{x - (- 4)} = \frac{y + 3}{x + 4}$

So, $\frac{d y}{d x} = 2 (\frac{y + 3}{x + 4})$

$\Rightarrow \frac{d y}{y + 3} = \frac{2}{x + 4} d x$

Integrating both sides,

$\begin{array}{l} \int \frac{d y}{y + 3} = \int \frac{2}{x + 4} d x \\ \Rightarrow l o g | y + 3 | = 2 l o g | x + 4 | + l o g | c | \\ \Rightarrow l o g | y + 3 | = l o g {(x + 4)}^{2} + l o g | c | \\ \Rightarrow l o g | y + 3 | = l o g | c {(x + 4)}^{2} | \\ \Rightarrow y + 3 = c_{1} {(x + 4)}^{2}, w h e r e, c_{1} = \pm c \end{array}$

Since, the curve passes through (-2,1) we get,

$\begin{array}{l} y = 1, a t, x = - 2 \\ \Rightarrow 1 + 3 = c {(- 2 + 4)}^{2} \\ \Rightarrow 4 = c * 4 \\ \Rightarrow c = 1 \end{array}$

$∴$ The equation of the curve is $y + 3 = {(x + 4)}^{2}$

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