Finding the tangent to a point on the curve

Question

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?

Answer 1

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 $y=\pm 2x$.

Answer 2

Note there is a faster way, which shows there are two solutions to this problem.

It amounts to finding the tangents which can be drawn from the origin to the parabola.

Now, a (non-vertical) straight line through the origin has equation $y=tx$ for some $t\in\mathbf R$, and such a line is tangent to the parabola if and only if the equation for the abscissæ of the intersection points: $$ tx=x^2+1\iff x^2-tx+1=0 $$ has a *double root, i.e. if and only if the discriminant of this equation: $ \Delta = t^2-4 \;$ is $0$. This $t=\pm2$, and rthe equations of the tangent lines from the origin are $$y=\pm2x. $$ Furthermore, the abscissæ of the points of contact with the parabola are the double roots: $x=\pm 1$.