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Show that the equation of the tangent to y=2x²-1 at the point where x=a, is 4ax-y=2a²+1

User Gorcyn
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Final answer:

To find the equation of the tangent to the curve y = 2x² - 1 at the point where x = a, you can find the slope of the tangent line using the derivative. The equation of the tangent line can be written in the form 4ax - y = 2a² + 1.

Step-by-step explanation:

To find the equation of the tangent to the curve y = 2x² - 1 at the point (a, 2a² - 1), we need to find the slope of the tangent line at that point. The slope of a curve at a given point can be found using the derivative. So, let's first take the derivative of y = 2x² - 1 with respect to x:

y' = 4x

Now, substitute the x-coordinate a into the derivative to find the slope at the point (a, 2a² - 1):

slope = 4a

Now, we have the slope and a point (a, 2a² - 1). We can use the point-slope form of a linear equation to find the equation of the tangent line:

y - (2a² - 1) = 4a(x - a)

Simplifying this equation gives us:

4ax - y = 2a² + 1

User Gkocjan
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