# 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) - Mathematics

Sum

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).

#### Solution

The slope of the line having points (xy) 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

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

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 12 Maths
Chapter 9 Differential Equations
Q 18 | Page 396
RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 71 | Page 147