The equation of the tangent to the curve y=sq | vedclass

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(A) Given the curve $y=\sqrt{9-2x^2}$.Let the point be $(x_1, y_1)$ such that $x_1 = y_1$.Substituting $y_1 = x_1$ into the curve equation: $x_1 = \sq

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$2x+y-3\sqrt{3}=0$

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(A) Given the curve $y=\sqrt{9-2x^2}$.Let the point be $(x_1, y_1)$ such that $x_1 = y_1$.Substituting $y_1 = x_1$ into the curve equation: $x_1 = \sqrt{9-2x_1^2}$.Squaring both sides: $x_1^2 = 9-2x_1^2 \Rightarrow 3x_1^2 = 9 \Rightarrow x_1^2 = 3 \Rightarrow x_1 = \pm\sqrt{3}$.Since $y = \sqrt{9-2x^2}$ must be positive,$y_1 = \sqrt{3}$. Thus,the point is $(\sqrt{3}, \sqrt{3})$.Now,find the slope $m = \frac{dy}{dx}$:$\frac{dy}{dx} = \frac{1}{2\sqrt{9-2x^2}} 	imes (-4x) = \frac{-2x}{\sqrt{9-2x^2}} = \frac{-2x}{y}$.At the point $(\sqrt{3}, \sqrt{3})$,the slope $m = \frac{-2(\sqrt{3})}{\sqrt{3}} = -2$.The equation of the tangent is $y - y_1 = m(x - x_1)$:$y - \sqrt{3} = -2(x - \sqrt{3})$$y - \sqrt{3} = -2x + 2\sqrt{3}$$2x + y - 3\sqrt{3} = 0$.

The equation of the tangent to the curve $y=\sqrt{9-2x^2}$ at the point where the ordinate and abscissa are equal is:

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