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Show that the equation of the tangent to y=2x21y=2 x^{2}-1 at the point where x = a, is 4axy=2a2+14 a x-y=2 a^{2}+1.

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Answered 2 years ago
Answered 2 years ago

Consider the curve y=2x21y=2x^2-1.

We have that dydx=4x\displaystyle\frac{dy}{dx}=4x and then when x=ax=a, dydx=4a\displaystyle\frac{dy}{dx}=4a.

We have also that when x=ax=a, y=2a21y=2a^2-1.

Hence, the equation of the tangent to the curve at the point (a,2a21)(a,2a^2-1) is y(2a21)xa=4a\displaystyle\frac{y-(2a^2-1)}{x-a}=4a which is the same as y2a2+1=4ax4a2y-2a^2+1=4ax-4a^2 or 4axy=2a2+14ax-y=2a^2+1.

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