The area (in sq. units) of the triangle forme | vedclass

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Question

(C) Given curve: $(\frac{x}{3})^n + (\frac{y}{4})^n = 2$. Differentiating with respect to $x$: $\frac{n}{3} (\frac{x}{3})^{n-1} + \frac{n}{4} (\frac{y

Vedclass · Accepted Answer

$\frac{50}{3}$

Vedclass · Answer

(C) Given curve: $(\frac{x}{3})^n + (\frac{y}{4})^n = 2$. Differentiating with respect to $x$: $\frac{n}{3} (\frac{x}{3})^{n-1} + \frac{n}{4} (\frac{y}{4})^{n-1} \frac{dy}{dx} = 0$. At point $(3,4)$, the slope of the tangent is: $\frac{dy}{dx} = -\frac{4}{3} (\frac{3/3}{4/4})^{n-1} = -\frac{4}{3}$. Equation of tangent at $(3,4)$: $y - 4 = -\frac{4}{3}(x - 3) \Rightarrow 4x + 3y = 24$. The $X$-intercept of the tangent is found by setting $y=0$, giving $4x = 24 \Rightarrow x = 6$. So, point $C$ is $(6,0)$. Slope of normal is $-\frac{1}{dy/dx} = \frac{3}{4}$. Equation of normal at $(3,4)$: $y - 4 = \frac{3}{4}(x - 3) \Rightarrow 3x - 4y + 7 = 0$. The $X$-intercept of the normal is found by setting $y=0$, giving $3x = -7 \Rightarrow x = -\frac{7}{3}$. So, point $B$ is $(-\frac{7}{3}, 0)$. The triangle is formed by vertices $A(3,4)$, $B(-\frac{7}{3}, 0)$, and $C(6,0)$. The base $BC = 6 - (-\frac{7}{3}) = 6 + \frac{7}{3} = \frac{25}{3}$. The height of the triangle is the $y$-coordinate of $A$, which is $4$. Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{25}{3} 	imes 4 = \frac{50}{3}$ sq. units.

The area (in sq. units) of the triangle formed by the tangent and normal drawn to the curve $(\frac{x}{3})^n+(\frac{y}{4})^n=2$ at $(3,4)$ and the $X$-axis is

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