Find an equation of the tangent line to the following curve (piriform) at the point:?

(1,1/2*2^(1/2))

4*y^2=x^3*(3-x)

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A few days ago

Anonymous

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4*y^2=x^3*(3-x)

4y^2 = 3x^3 – x^4

8y dy/dx = 9x^2 – 4x^3

dy/dx = (9x^2 – 4x^3) / 8y

The gradient at the point

(1,1/2*2^(1/2))

is:

(9 – 4) / 4sqrt(2)

= 5sqrt(2) / 8

The tangent is therefore:

y – (1/2)sqrt(2) = 5sqrt(2)(x – 1) / 8

8y – 4sqrt(2) = 5sqrt(2)x – 5sqrt(2)

8y – 5sqrt(2)x + sqrt(2) = 0.

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