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**Find the equation of tangent and normal to the curve at the given points on it.**

2x^{2} + 3y^{2} = 5 at (1, 1)

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#### Solution

Equation of the curve is 2x^{2} + 3y^{2} = 5

Differentiating w.r.t. x, we get

4x + 6y `*"dy"/"dx"` = 0

∴ `"dy"/"dx" = (- "4x")/"6y"`

∴ Slope of the tangent at (1, 1) is

`("dy"/"dx")_((1,1)` = `(-4(1))/(6(1)) = (-2)/3`

∴ Equation of tangent at (a, b) is

y - b = `("dy"/"dx")_(("a, b")` (x - a)

Here, (a, b) ≡ (1, 1)

∴ Equation of the tangent at (1, 1) is

(y - 1) = `(-2)/3`(x - 1)

∴ 3(y - 1) = - 2(x - 1)

∴ 3y - 3 = - 2x + 2

∴ 3y - 3 = - 2x + 2

∴ 2x + 3y - 5 = 0

Slope of the normal at (1, 1) is `(-1)/(("dy"/"dx")_((1,1)` `= 3/2`

∴ Equation of normal at (a, b) is

y - b = `(-1)/(("dy"/"dx")_(("a","b")` (x - a)

∴ Equation of the normal at (1, 1) is

(y - 1) = `3/2`(x - 1)

∴ 2y - 2 = 3x - 3

∴ 3x - 2y - 1 = 0

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