The normal at the point (1,1) on the curve 2y | vedclass

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Question

(A) The equation of the given curve is $2y + x^{2} = 3$. Differentiating with respect to $x$,we get: $2 \frac{dy}{dx} + 2x = 0$ $\Rightarrow \frac{dy}

Vedclass · Accepted Answer

$x - y = 0$

Vedclass · Answer

(A) The equation of the given curve is $2y + x^{2} = 3$. Differentiating with respect to $x$,we get: $2 \frac{dy}{dx} + 2x = 0$ $\Rightarrow \frac{dy}{dx} = -x$. At the point $(1, 1)$,the slope of the tangent is: $\left. \frac{dy}{dx} ight|_{(1, 1)} = -1$. The slope of the normal at $(1, 1)$ is: $m = \frac{-1}{	ext{slope of tangent}} = \frac{-1}{-1} = 1$. The equation of the normal at $(1, 1)$ is given by: $y - y_{1} = m(x - x_{1})$ $y - 1 = 1(x - 1)$ $y - 1 = x - 1$ $x - y = 0$. Thus,the correct option is $A$.

The normal at the point $(1,1)$ on the curve $2y + x^{2} = 3$ is

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