Math Help - Finding a Tangent Line without Calculus

1. Finding a Tangent Line without Calculus

I hope this is in the right place, I'm not in a hurry, just curious.

How can I find an equation for a line tangent to a point on a parabola without using calculus?

I just started playing with this this morning

The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)

Using calculus I found the equation to be y = 4x -18

How can I do this without calculus?

2. Hello, Ranger SVO!

You're curious . . . good for you!

While playing with parabolas in college (centuries ago),
. . I "discovered" some facts that might help you.

The equation I'm using is $y \:= \:x^2 - 4x - 2$
and I'm looking for the equation of the tangent line at point $(4,\,-2)$

Using calculus I found the equation to be: $y \:= \:4x -18$

How can I do this without calculus?
Code:
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|       /
*      |      *P(x,y)
*    |    */
---------***---o----------
|   /↑
|  /(½x,0)
| /
|/
o ←(0,-y)
/|

Given the parabola: . $y \:=\:ax^2$, the tangent at point $P(x,y)$
. . has an x-intercept of $\left(\frac{1}{2}x,\,0\right)$ and a y-intercept of $(0,\,-y)$

The tangent at $P(x,y)$ always has a point directly below the vertex
. . with the same vertical displacement (only downward).

Your parabola $y \:=\:x^2-4x-2$ has its vertex at $(2,-6).$

Your point $P(4,-2)$ is 4 units above the level of the vertex.
Hence, it has a point 4 units directly below the vertex: $Q(2,-10).$

Now you can write the equation through $P$ and $Q.$

3. Originally Posted by Ranger SVO
I hope this is in the right place, I'm not in a hurry, just curious.

How can I find an equation for a line tangent to a point on a parabola without using calculus?

I just started playing with this this morning

The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)

Using calculus I found the equation to be y = 4x -18

How can I do this without calculus?
I have another approach.
It does not always work.
But it does sometimes.

In your example, the tangent line is the line that only intersects the parabola at one point. Thus, you need to find an equation of a line that one intersects at one point, i.e. solution set is only one element. That means you need the discrimanant to be zero.

4. Soroban, I like your explination. But first, at my age curiousity is the only thing that keeps me from vegetating.

Here is what I've been working on. If I draw a rectangle using the vertex and the point where I want the tagent line, I can see the secant line bisecting the rectangle. I can easily notice that the slope of the tangent is twice that of the secant.

Is this always true? If so why?

Perfect Hacker, does that mean that I can use the quadratic equation to find the slope of a line tangent to some point.

5. Hello again, Ranger SVO!

Here is what I've been working on.
If I draw a rectangle using the vertex and the point where I want the tangent line,
I can see the secant line bisecting the rectangle.
I can easily notice that the slope of the tangent is twice that of the secant.

Is this always true? If so why?

Given the parabola: $y \:=\:ax^2$ and any point on it: $P(p,\,ap^2)$

The slope of the secant $OP$ is: . $m_s \:=\:\frac{ap^2 - 0}{p - 0} \:=\:ap$

The derivative is: . $y' \:=\:2ax$
. . Hence, the slope of the tangent at $P$ is: $m_t\:=\:2ap$

Therefore, the slope of the tangent is always twice the slope of the secant.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you're really bored, verify these claims.

1) On the parabola $y \:=\:ax^2$, pick any two points: . $P(p,\,ap^2)$ and $Q(q,\,aq^2)$

Draw the chord $PQ.$

Locate the tangent parallel to $PQ.$

It is found at: . $x \:=\:\frac{p+q}{2}$ . . . halfway between $p$ and $q.$

2) On the parabola $y \:=\:ax^2$, pick any three points: . $P(p,\,ap^2),\;Q(q,\,aq^2),\;R(r,\,ar^2)$
Find the area of $\Delta PQR.$

Consider the tangents to the parabola at $P,\,Q,\,R.$
. . They will intersect at $A,\,B,\,C.$
Find the area of $\Delta ABC.$

We find that: . $(\text{area }\Delta ABC) \:=\:\frac{1}{2}(\text{area }\Delta PQR)$

6. y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
We want a tangent line at $(4,-2)$.
To do that I assume you are familar with the point-slope formula:
$y-y_0=m(x-x_0)$
In this case,
$(x_0,y_0)=(4,-2)$.

Thus, the equation of the line (tangent) at that point is determined by,
$y+3=m(x-4)$ that means ( $y=m(x-4)-3$)
Where,
$m$
Is the slope to be determined.

As I said we need to intersection to be a single point.
That is,
$\left\{ \begin{array}{c}y=x^2-4x-2\\y=m(x-4)-3 \end{array} \right\}$
We want this to has a single solution.
Equate,
$x^2-4x-2=m(x-4)-3$
$x^2-4x-2=mx-4m-3$
$x^2+x(-4-m)+(1+4m)=0$
To have a single real solution we need the discrimanant to be zero,
$(-4-m)^2-4(1)(1+4m)=0$
Solve to get the value of $m$.