The equation of the normal to the curve 2x^{2 | vedclass

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(D) Given the equation of the curve is $2x^{2} + y^{2} = 12$. Differentiating with respect to $x$,we get $4x + 2y \frac{dy}{dx} = 0$. Thus,$\frac{dy}{

Vedclass · Accepted Answer

$x - 2y + 2 = 0$

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(D) Given the equation of the curve is $2x^{2} + y^{2} = 12$. Differentiating with respect to $x$,we get $4x + 2y \frac{dy}{dx} = 0$. Thus,$\frac{dy}{dx} = -\frac{2x}{y}$. The slope of the tangent at $(2, 2)$ is $m_{t} = -\frac{2(2)}{2} = -2$. The slope of the normal at $(2, 2)$ is $m_{n} = -\frac{1}{m_{t}} = -\frac{1}{-2} = \frac{1}{2}$. The equation of the normal at $(2, 2)$ is given by $y - y_{1} = m_{n}(x - x_{1})$. Substituting the values,$y - 2 = \frac{1}{2}(x - 2)$. Multiplying by $2$,we get $2y - 4 = x - 2$. Rearranging the terms,$x - 2y + 2 = 0$. Therefore,the correct option is $D$.

The equation of the normal to the curve $2x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

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