Write the equation of tangent at (1, 1) on the curve $2 x^{2} + 3 y^{2} = 5$ .

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Solution

The slope of the tangent to the curve $y = f (x)$ at a point $(x, y)$ is given by $\frac{d y}{d x}$
$2 x^{2} + 3 y^{2} = 5 ⟹ 4 x + 6 y \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = \frac{- 2 x}{3 y}$
AT $(1, 1), \frac{d y}{d x} = \frac{- 2}{3}$
Equation of line passing through $(1, 1)$ and having a slope of $\frac{- 2}{3}$ is given as
$y - 1 = \frac{- 2}{3} (x - 1) ⟹ 2 x + 3 y = 5$

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