The normal at the point (1, 1) on the curve 2y + x2 = 3 is _____________ .

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x − y = 0 	ext { Given }: 2y + x^2 = 3 \Rightarrow 2dy/dx + 2x = 0 \Rightarrow dy/dx = - 2x/2 = - x 	ext { Slope of the tangent } = \left( dy/dx ight)_\left( 1, 1 ight) =-1 	ext { Slope of the normal },m=\frac{- 1}{	ext { Slope of the tangent }}=- 1/- 1=1 	ext { Now }, \left( x_1 , y_1 ight) = \left( 1, 1 ight) 	herefore 	ext { Equation of the normal } = y - y_1 = m \left( x - x_1 ight) \Rightarrow y - 1 = 1 \left( x - 1 ight) \Rightarrow x - y = 0

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