The tangent and normal to the ellipse 3x^2 + | vedclass

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Question

(B) Given the ellipse equation: $3x^2 + 5y^2 = 32$. Differentiating with respect to $x$: $6x + 10y \frac{dy}{dx} = 0$,which gives $\frac{dy}{dx} = -\f

Vedclass · Accepted Answer

$\frac{68}{15}$

Vedclass · Answer

(B) Given the ellipse equation: $3x^2 + 5y^2 = 32$. Differentiating with respect to $x$: $6x + 10y \frac{dy}{dx} = 0$,which gives $\frac{dy}{dx} = -\frac{3x}{5y}$. At point $P(2, 2)$,the slope of the tangent is $m = -\frac{3(2)}{5(2)} = -\frac{3}{5}$. The equation of the tangent at $P(2, 2)$ is $y - 2 = -\frac{3}{5}(x - 2)$. Setting $y = 0$ to find $Q$,we get $-2 = -\frac{3}{5}(x - 2)$ $\Rightarrow x - 2 = \frac{10}{3}$ $\Rightarrow x = \frac{16}{3}$. Thus,$Q = (\frac{16}{3}, 0)$. The slope of the normal at $P(2, 2)$ is $m' = -\frac{1}{m} = \frac{5}{3}$. The equation of the normal at $P(2, 2)$ is $y - 2 = \frac{5}{3}(x - 2)$. Setting $y = 0$ to find $R$,we get $-2 = \frac{5}{3}(x - 2)$ $\Rightarrow x - 2 = -\frac{6}{5}$ $\Rightarrow x = \frac{4}{5}$. Thus,$R = (\frac{4}{5}, 0)$. The area of triangle $PQR$ with base $QR$ on the $x$-axis and height equal to the $y$-coordinate of $P$ is: Area $= \frac{1}{2} 	imes |x_Q - x_R| 	imes |y_P| = \frac{1}{2} 	imes |\frac{16}{3} - \frac{4}{5}| 	imes 2 = |\frac{80 - 12}{15}| = \frac{68}{15}$ sq. units.

The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x$-axis at $Q$ and $R$,respectively. Then the area (in sq. units) of the triangle $PQR$ is

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