## Related questions with answers

Question

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

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

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

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

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

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

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