The point on the curve y=x^2+4x+3 which is cl | vedclass

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(B) Let the point on the curve be $P(x, y)$ where $y = x^2 + 4x + 3$. The line is $3x - y + 2 = 0$. For the point to be closest to the line,the tangen

Vedclass · Accepted Answer

$\left(-\frac{1}{2}, \frac{5}{4}\right)$

Vedclass · Answer

(B) Let the point on the curve be $P(x, y)$ where $y = x^2 + 4x + 3$. The line is $3x - y + 2 = 0$. For the point to be closest to the line,the tangent at that point must be parallel to the given line. The slope of the tangent is $\frac{dy}{dx} = 2x + 4$. Since the tangent is parallel to $y = 3x + 2$,its slope must be $3$. $2x + 4 = 3 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}$. Now,find the $y$-coordinate by substituting $x = -\frac{1}{2}$ into the curve equation: $y = \left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{5}{4}$. Thus,the point is $\left(-\frac{1}{2}, \frac{5}{4}\right)$.

The point on the curve $y=x^2+4x+3$ which is closest to the line $y=3x+2$ is

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