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

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Question

(D) Given the curve equation: $2x^{2} + 3y^{2} - 5 = 0$. Differentiating both sides with respect to $x$: $4x + 6y \frac{dy}{dx} = 0$ $\frac{dy}{dx} =

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$3x - 2y - 1 = 0$

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(D) Given the curve equation: $2x^{2} + 3y^{2} - 5 = 0$. Differentiating both sides with respect to $x$: $4x + 6y \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y}$. At the point $P(1, 1)$,the slope of the tangent $(m_{t})$ is: $m_{t} = -\frac{2(1)}{3(1)} = -\frac{2}{3}$. The slope of the normal $(m_{n})$ is the negative reciprocal of the tangent slope: $m_{n} = -\frac{1}{m_{t}} = -\frac{1}{-2/3} = \frac{3}{2}$. The equation of the normal at $(x_{1}, y_{1}) = (1, 1)$ is given by: $y - y_{1} = m_{n}(x - x_{1})$ $y - 1 = \frac{3}{2}(x - 1)$ $2(y - 1) = 3(x - 1)$ $2y - 2 = 3x - 3$ $3x - 2y - 1 = 0$.

The equation of the normal to the curve $2x^{2} + 3y^{2} - 5 = 0$ at the point $P(1, 1)$ is:

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