Find the equation for a line tangent to an equation with a point

Question

Say we have the equation $y=e^{x^2}$. How do I find the equation of a line tangent to $y$ that also passes through $(1,0)$, or any point?

So far I only know the derivative of $y$ which is $2xe^{x^2}$.

Answer 1

I have a solution a little different from DavidP's

Consider the line containing $(1,0)$ and $\left(t,e^{t^2}\right)$. The slope of this line is $\displaystyle\frac{e^{t^2}}{t-1}$. Now, note that $$\frac{d}{dt}\frac{e^{t^2}}{t-1}=\frac{\left(2t^2-2t-1\right)e^{t^2}}{(t-1)^2}$$

This expression is equal to zero when $t=\displaystyle\frac{1\pm\sqrt{3}}{2}$, so the tangent lines we are looking for intersect the graph $y=e^{x^2}$ at $x=\displaystyle\frac{1\pm\sqrt{3}}{2}$.

I figured out this solution by drawing pictures.

The black point represents $(1,0)$. The blue point represents the point where the tangent line intersects the curve. The green point represents a random point on the curve.

Imagine the green point starting from the left of the blue point, moving to the right. The slope of the orange line increases, until the green point reaches the blue point, at which point the slope starts decreasing. So, if we write the slope of the orange line as a function of the $x$-coordinate of the green point, the function's derivative will equal zero at the $x$-coordinate of the blue point.

Answer 2

First we need to know where the line is tangent. Once you have your point(s) $(x_i,f(x_i))$ the slopes are $f'(x_i)$.

The line passes through $(1,0)$ and another point(s) $(x,e^{x^2})$. So on one hand,

$$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{e^{x^2}}{x-1}$$

However, we also know $m=f'(x) = 2xe^{x^2}$. Therefore

$$\dfrac{e^{x^2}}{x-1} = 2xe^{x^2} \iff 2x^2-2x-1=0$$

As this has two unique solutions, you will find two different points on the graph with the required properties, and so two different lines. An exact presentation will be a little messy.