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Question
Find the equation to the tangent at (0, 0) on the curve y = 4x2 – 2x3
Solution
y = 4x2 – 2x3
`dy/dx = 8x - 6x^2`
Let P(x1, y1) be the point of contact.
∴ Slope of the tangents = `8x_1 - 6x_1^2`
Tangent equation: y – y1 = `(8x_1 - 6x_1^2)(x - x_1)`
∵ The tangent passes through the origin
`y_1 = (8x_1 - 6x_1^2)x_1`
Now, `y_1 = 4x_1^2 - 2x_1^3`
= `8x_1^2 - 6x_1^3`
`\implies` 4 – 2x1 = 8 – 6x1
`\implies` 4x1 = 4
`\implies` x1 = 1
`\implies` y1 = 2
The point on the curve is (1, 2)
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