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synkk
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The point P(2, 8) lies on the parabola C with equation y2=4ax. Find
a the value of a,
b an equation of the tangent to C at P.
the value of a is 8, so y^2 = 32x
when finding the tangent [itex] y = 4\sqrt{2} x^{\frac{1}{2}} [/itex] so at P [itex] \frac{dy}{dx} = \frac{2\sqrt{2}}{\sqrt{2}} [/itex] so the gradient of the tangent is 2, but why isn't it also -2? as when differentiating y^2 = 32x, you get also [itex] y = -4\sqrt{2} x^{\frac{1}{2}} [/itex] Could anyone explain why they have only used the positive value?
a the value of a,
b an equation of the tangent to C at P.
the value of a is 8, so y^2 = 32x
when finding the tangent [itex] y = 4\sqrt{2} x^{\frac{1}{2}} [/itex] so at P [itex] \frac{dy}{dx} = \frac{2\sqrt{2}}{\sqrt{2}} [/itex] so the gradient of the tangent is 2, but why isn't it also -2? as when differentiating y^2 = 32x, you get also [itex] y = -4\sqrt{2} x^{\frac{1}{2}} [/itex] Could anyone explain why they have only used the positive value?
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