# Thread: [SOLVED] Tangent Line/Secant Line To A Parabola

1. ## [SOLVED] Tangent Line/Secant Line To A Parabola

You can find the slope of the tangent line to a parabola $\displaystyle y=x^2$ at a point $\displaystyle P(a,a^2)$ using the following method.

(a) Find an expression, in terms of x and a, for the slope of a secant line from $\displaystyle P$ to another point $\displaystyle Q(x,x^2)$ on the parabola.

(b) Simplify this slope assuming $\displaystyle x \not= a$. What number, in terms of $\displaystyle a$, does this quantity approach as $\displaystyle x$ approaches $\displaystyle a$?
Well for part (a) since the equation of a slope is $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ then the expression should be $\displaystyle \frac{a^2-x^2}{a-x}$ right?

I don't understand part (b) at all.

2. Hello,

They're asking you to what value the quotient tends when x will be about to equal a. It's called the limit of the expression when x tends to a.

For this thing, you can factorize (a²-x²) with the (a-b)(a+b)=a²-b² formula. And as $\displaystyle x \neq a$, $\displaystyle a-x \neq 0$, you can simplify ;-)

If you don't understand the notion of limit, it's like replacing x by a.

3. Originally Posted by Moo
Hello,

They're asking you to what value the quotient tends when x will be about to equal a. It's called the limit of the expression when x tends to a.

For this thing, you can factorize (a²-x²) with the (a-b)(a+b)=a²-b² formula. And as $\displaystyle x \neq a$, $\displaystyle a-x \neq 0$, you can simplify ;-)

If you don't understand the notion of limit, it's like replacing x by a.
Ok but for the second part it said to simplify the slope instead of $\displaystyle x \neq a$. How would I go by simplifying the slope?

Edit - nevermind. I got the slope to be simplified to $\displaystyle a+x$. But how would go you go about to find the "the limit of expression" when $\displaystyle x$ approaches $\displaystyle a$.

4. When x approaches a, you can say that x+a approaches a+a, which is 2a. And this is what you're asked to get :-)

5. Originally Posted by Moo
When x approaches a, you can say that x+a approaches a+a, which is 2a. And this is what you're asked to get :-)
Ok...I understand now. Thx alot.