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

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(D) Given curve is $y=\sqrt{9-2x^2}$.If ordinate and abscissa are equal,then $y=x$.Substituting $y=x$ in the curve equation: $x^2 = 9 - 2x^2 \Rightarr

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

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(D) Given curve is $y=\sqrt{9-2x^2}$.If ordinate and abscissa are equal,then $y=x$.Substituting $y=x$ in the curve equation: $x^2 = 9 - 2x^2 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \pm \sqrt{3}$.Since $y = \sqrt{9-2x^2}$,$y$ must be positive. Thus,$x$ must be $\sqrt{3}$ because if $x = -\sqrt{3}$,then $y = \sqrt{3}$,but $x 
eq y$.So,the point of contact is $(\sqrt{3}, \sqrt{3})$.Differentiating $y^2 = 9 - 2x^2$ with respect to $x$: $2y \frac{dy}{dx} = -4x \Rightarrow \frac{dy}{dx} = -\frac{2x}{y}$.At $(\sqrt{3}, \sqrt{3})$,the slope $m = -\frac{2(\sqrt{3})}{\sqrt{3}} = -2$.The equation of the tangent is $y - \sqrt{3} = -2(x - \sqrt{3})$.$y - \sqrt{3} = -2x + 2\sqrt{3} \Rightarrow 2x + y - 3\sqrt{3} = 0$.

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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