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.