P(5,2) is a point on the curve y=f(x) and fra | vedclass

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(C) Given point $P(5,2)$ and slope of the tangent $m_t = \frac{7}{2}$.Equation of the tangent at $P(5,2)$ is $y - 2 = \frac{7}{2}(x - 5) \implies 2y -

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$\frac{53}{7}$

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(C) Given point $P(5,2)$ and slope of the tangent $m_t = \frac{7}{2}$.Equation of the tangent at $P(5,2)$ is $y - 2 = \frac{7}{2}(x - 5) \implies 2y - 4 = 7x - 35 \implies 7x - 2y = 31$.The $x$-intercept of the tangent is found by setting $y=0$: $7x = 31 \implies x = \frac{31}{7}$. So, the tangent meets the $x$-axis at $A(\frac{31}{7}, 0)$.The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{2}{7}$.Equation of the normal at $P(5,2)$ is $y - 2 = -\frac{2}{7}(x - 5) \implies 7y - 14 = -2x + 10 \implies 2x + 7y = 24$.The $x$-intercept of the normal is found by setting $y=0$: $2x = 24 \implies x = 12$. So, the normal meets the $x$-axis at $B(12, 0)$.The triangle is formed by points $P(5,2)$, $A(\frac{31}{7}, 0)$, and $B(12, 0)$.The base of the triangle on the $x$-axis is $|12 - \frac{31}{7}| = |\frac{84 - 31}{7}| = \frac{53}{7}$.The height of the triangle is the $y$-coordinate of $P$, which is $2$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{53}{7} 	imes 2 = \frac{53}{7}$ sq. units.

$P(5,2)$ is a point on the curve $y=f(x)$ and $\frac{7}{2}$ is the slope of the tangent to the curve at $P$. The area of the triangle (in sq. units) formed by the tangent and the normal to the curve at $P$ with the $x$-axis is:

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