[Math] Finding the tangent to a point on the curve

curvesderivativestangent line

I was able to figure out the answer to the following question but I didn't understand how the process worked.

Question:

The tangent at the point P on the curve $ y=x^2+1 $ passes through the origin. Find the equation to the tangent.

My first impression was to find the gradient of the tangent by finding the derivative of the curve.

$\frac{dy}{dx}= 2x
$

I let the point P = a,b

p(a,b)

Hence

$\frac{dy}{dx}= 2a
$

I then constructed the equation of the tangent.

$y=(2a)x+c $

I know $C=0$ as the tangent passes through (0,0)

Hence

$y=(2a)x$

I found the derivative ( I don't understand why) of $\frac{dy}{dx}= 2a
$ to give the gradient of 2 completing the equation of the tangent to the curve:

$y=2x+c $

My question is how does finding the second derivative of the curve result in the gradient of the tangent?

Best Answer

Your work is ok, you can now find the point $P(a,b)$ by the fact that it must belongs both to the tangent and to the parabola, notably

$$y=(2a)x \implies b=2a^2$$

and thus

$$y=x^2+1 \implies 2a^2=a^2+1 \implies a=\pm1 \quad b=2$$

Now verify which one is good or if both work.

By symmetry it easy verify that both values work for $P(\pm 1,2)$ with tangents $y=\pm 2x$.

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