Calculus problem of finding the equation of a line.

Question

Find the equation of a line that passes through the origin, with positive slope, and its tangent to the parabola given by :$ y = x^2 - 2x + 2$

My approach to this problem was to differentiate the equation of the parabola, so I can et an expression, that determines the tangent line anywhere on the parabola. I then got $2x - 2$, and I replaced x with a point (p), which is the second point of the line, where the first point is the origin $(0,0)$, and the second point being (p) which is tangent to the parabola, I then used $2p - 2$ as the slope of , my line, and the y intercept of the line is zero, because the line goes through the origin. The equation of the then becomes : $y = x (2p-2)$, and now I set the line and the parabola equal to each other, and at that point is where I get stuck, because I cant solve for $(p)$. I am pretty sure that my approach is wrong, and this is why I needed help solving this problem.

Could somebody, who solves this problem, provide a full solution below? Thank You !

Answer 1

A generic line is given by $y = ax + b$.

If it passes in $(x=0,y=0)$ you can put this value in your equation and find $0 = a0 + b$, thus $b = 0$, and your line is $y =ax$.

The common points of the line and the parabola are given by $ax = y = x^2 -2x +2$ thus $x$ must solve $ax = x^2 -2x +2$. Rewrite it like $x^2 + x(-a-2) +2 = 0$.

This is an equation of degree 2. In order to be tangent, you need to find exactly one solution $x$ (not $0$, or the line doesn't intersect your parabola, and not $2$, or it passes through the parabola and it's not tangent).

You should know that such an equation has exactly one solution when the discriminant $\Delta = b^2 - 4ac$ (not the same $a$ of the equation, here it's just a symbol!) it's equal to $0$.

In our case we must solve $\Delta = (-a-2)^2 -4\cdot 1 \cdot 2 = 0$.

Rewrite as $a^2 +4a -4 = 0$. This is again an equation of degree two, we can solve with the well know formula and get $a_1$ and $a_2$. But only one of them is positive, the other is negative. Since you are looking for positive slope, you must take the positive one!

Answer 2

At the point $x=x_0$ the slope of the line,as you said, is :

$$2x_0-2$$

And so the line is of the form:

$$y=(2x_0-2)x+b$$

What is $b$?

You did say it passes through the origin where $x=0$ and $y=0$. So with some basic algebra we can come up with:

$$b=0$$

$$y=(2x_0-2)x$$

Again, this you correctly executed.Now all that is left is to conclude what $x_0$ and the corresponding slope(s) may be. How do we do this?

We incorporate our final bit out information. That our line must pass through the point:

$$(x_0,x_0^2-2x_0+2)$$

Because it is tangent to our function at $x=x_0$.

And we get,

$$x_0^2-2x_0+2=(2x_0-2)x_0$$

$$x_0=\sqrt{2},x_0=-\sqrt{2}$$

$$y=(2\sqrt{2}-2)x,y=(-2\sqrt{2}-2)x$$

But you only want the one with a positive slope which is:

$$y=(2\sqrt{2}-2)x$$

So really the trick is to plug in $x=p$ and $y=p^2-2p+2$ into your linear equation.