At any point (x, y) of a curve,the slope of t | vedclass

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(A) Let $(x, y)$ be any point on the curve.The slope $(m_1)$ of the line segment joining $(x, y)$ and $(-4, -3)$ is given by $m_1 = \frac{y - (-3)}{x

Vedclass · Accepted Answer

$y+3=(x+4)^2$

Vedclass · Answer

(A) Let $(x, y)$ be any point on the curve.The slope $(m_1)$ of the line segment joining $(x, y)$ and $(-4, -3)$ is given by $m_1 = \frac{y - (-3)}{x - (-4)} = \frac{y+3}{x+4}$.The slope $(m_2)$ of the tangent to the curve at $(x, y)$ is $\frac{dy}{dx}$.According to the problem,$m_2 = 2m_1$.Therefore,$\frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right)$.Separating the variables,we get $\frac{dy}{y+3} = \frac{2 dx}{x+4}$.Integrating both sides,$\int \frac{dy}{y+3} = 2 \int \frac{dx}{x+4}$.This gives $\ln|y+3| = 2 \ln|x+4| + C_1$,which can be written as $\ln|y+3| = \ln|x+4|^2 + \ln C$.Thus,$y+3 = C(x+4)^2$.Since the curve passes through $(-2, 1)$,we substitute $x = -2$ and $y = 1$:$1+3 = C(-2+4)^2 \Rightarrow 4 = C(2)^2 \Rightarrow 4 = 4C \Rightarrow C = 1$.Substituting $C=1$ into the general equation,we get $y+3 = (x+4)^2$.

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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