Consider a curve y=y(x) in the first quadrant | vedclass

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(C) Given that $A_{1} = 2A_{2}$.From the graph,the total area of the rectangle formed by the coordinates $(x, y)$ is $A_{1} + A_{2} = xy - (4 	imes 2

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$(10, 4)$

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(C) Given that $A_{1} = 2A_{2}$.From the graph,the total area of the rectangle formed by the coordinates $(x, y)$ is $A_{1} + A_{2} = xy - (4 	imes 2) = xy - 8$.Since $A_{1} = 2A_{2}$,we have $A_{1} + \frac{1}{2}A_{1} = xy - 8$,which implies $\frac{3}{2}A_{1} = xy - 8$,so $A_{1} = \frac{2}{3}xy - \frac{16}{3}$.Also,$A_{1} = \int_{4}^{x} y \, dx$. Differentiating with respect to $x$ using the Leibniz rule:$y = \frac{2}{3}(y + x \frac{dy}{dx}) \implies 3y = 2y + 2x \frac{dy}{dx} \implies y = 2x \frac{dy}{dx}$.Separating variables: $\int \frac{dy}{y} = \int \frac{dx}{2x} \implies \ln y = \frac{1}{2} \ln x + C \implies y^2 = cx$.Since the curve passes through $(4, 2)$,$2^2 = c(4) \implies c = 1$. Thus,$y^2 = x$.The line is $2x - 12y = 15$,or $y = \frac{1}{6}x - \frac{15}{12}$. Its slope is $m = \frac{1}{6}$.The normal is perpendicular to this line,so its slope is $m_n = -6$.For $y^2 = x$,$2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$.Setting $\frac{dy}{dx} = -\frac{1}{m_n} = \frac{1}{6}$,we get $\frac{1}{2y} = \frac{1}{6} \implies y = 3$. Since $y^2 = x$,$x = 9$.The point of contact is $(9, 3)$. The equation of the normal is $y - 3 = -6(x - 9) \implies y = -6x + 57$.Checking the options: For $(10, 4)$,$4 = -6(10) + 57 = -3$,which is false. Thus,the normal does not pass through $(10, 4)$.

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

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