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प्रश्न
Find the equation of tangent and normal to the curve at the given points on it.
x2 + y2 + xy = 3 at (1, 1)
उत्तर
Equation of the curve is x2 + xy + y2 = 3
Differentiating w.r.t. x, we get
2x + x `*"dy"/"dx" + "y" + 2"y" "dy"/"dx"` = 0
∴ (2x + y) + (x + 2y) `"dy"/"dx"` = 0
∴ `"dy"/"dx" = (-(2"x" + "y"))/("x" + "2y")`
∴ Slope of the tangent at (1, 1) is
`("dy"/"dx")_((1,1)` = `(-(2 + 1))/(1 + 2) = -1`
∴ 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) = -1 (x - 1)
∴ (y - 1) = - x + 1
∴ x + y - 2 = 0
Slope of the normal at (1, 1) is `(-1)/(("dy"/"dx")_((1,1)` = 1
∴ 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) = 1 (x - 1)
∴ x - y = 0
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