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).

Answer 1

The slope of the line having points (x, y) and (−4, −3) is given by

\[\frac{y + 3}{x + 4}\]

According to the question,

\[\frac{dy}{dx} = 2\left( \frac{y + 3}{x + 4} \right)\]

\[\Rightarrow \frac{1}{y + 3}dy = \frac{2}{x + 4}dx\]

Integrating both sides, we get

\[\int\frac{1}{y + 3}dy = 2\int\frac{1}{x + 4}dx\]

\[ \Rightarrow \log \left| y + 3 \right| = 2\log \left| x + 4 \right| + \log C\]

\[ \Rightarrow \log \left| y + 3 \right| = \log \left| C \left( x + 4 \right)^2 \right|\]

\[ \Rightarrow y + 3 = C \left( x + 4 \right)^2 \]

Since the curve passes through (-2, 1), it satisfies the equation of the curve.

\[ \therefore 1 + 3 = C \left( - 2 + 4 \right)^2 \]

\[ \Rightarrow C = 1\]

Putting the value of `C` in the equation of the curve, we get

\[y + 3 = \left( x + 4 \right)^2\]