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प्रश्न
Find the equation of tangent and normal to the curve at the given points on it.
2x2 + 3y2 = 5 at (1, 1)
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उत्तर
Equation of the curve is 2x2 + 3y2 = 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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