Tangents Show that the tangents to the curve y^2=4 p x from any point on the line x=-p are perpe

Tangents Show that the tangents to the curve y^2=4 p x from...

Key Concepts

Parabola

A parabola is a conic section defined as the set of all points that are equidistant from a fixed point called the focus and a fixed line called the directrix. Its geometric properties, symmetry, and standard equations form the basis for analyzing curves and determining tangent properties.

Tangent to a Parabola

A tangent line to a parabola is a straight line that touches the curve at exactly one point. The equation of the tangent can be derived using various methods, such as differentiation or by employing a standard tangent form for the parabola. This concept is crucial for understanding how lines interact with the curve.

External Point Tangency

Drawing tangent lines from an external point to a conic section involves determining the lines that simultaneously satisfy the condition of being a tangent and passing through the given external point. This concept is fundamental in analytic geometry and helps in solving problems involving touch points and contact chords.

Perpendicular Lines

Two lines are perpendicular if the product of their slopes is -1. This condition is used to determine the relative orientation of the tangent lines and is essential when proving that certain tangent lines to a curve are orthogonal, as it provides the mathematical criterion for their perpendicularity.

Transcript

00:01 We want to show that the tangents to the curve, y squared is equal to 4 px, from any point on the line, x is equal to negative p r perpendicular.

00:14 So let's go ahead and kind of sketch a graph of what they're wanting us to do, to maybe give us some intuition on how to begin.

00:23 So let's say we have this horizontal parabola here, and this will be our line x is equal to negative p.

00:43 So if we were to look at the points tangent that include this to our curve, well, it looks something kind of like this here, for one, and our other line would look something kind of like this.

01:05 So what we want to do is show that these two lines here are perpendicular to each other.

01:13 So let's go ahead.

01:16 And well, anytime we're dealing with tangent lines, we probably want to find the derivative of our function.

01:21 So let's go ahead and do that really quickly.

01:23 So the derivative of y squared is equal to 4 px with respect to x would be 2yy prime.

01:32 Because remember, we'd have to use implicit differentiation since y is a function that depends on x.

01:38 And then the right hand side would just be 4.

01:41 P.

01:42 Now, solving for y, or y prime, we'd get y prime is equal to 2p over y.

01:53 Now, this will be equal to, well, the slope of these tangents.

02:01 And you might recall that any point along this line, x is equal to negative p, would be negative p t, where t is just some value.

02:14 All right.

02:15 So, so let's go ahead and do this.

02:19 So the slope of a tangent line, or just a line, would be two points minus itself.

02:27 So our original point is going to be y and x, and then for our point along our line, x is equal to negative p, will that be t.

02:41 And then we'd have negative p for the x, but those just counts out with each other.

02:48 Now, the first thing i'm going to go ahead and do is multiply each side, because what we want to do is we want to find out what is y in this case.

03:00 Because once we figure out what y is, we can plug this in to our slope to see if the slopes are negative reciprocals of each other.

03:13 All right, so let's go ahead and multiply each side by y and x plus p.

03:18 And so doing that will give us 2p x plus p and over on the left we'll have y y minus t now x depends on y so let's go ahead and solve for x really quickly so we would need to divide this by 4p and doing that would tell us x is equal to y squared over 4 p so we can go ahead and plug that in right there so we have y squared over 4p plus p times 2p and then on the left -hand side we'll have y -squared minus y -t now let's go ahead and simplify our left -hand side a little bit so distributing the 2p we would end up with y -squared over 2 plus 2 p squared is equal to y -squared minus y t and now let's go ahead and move our work up top here so i'm going to move everything to the left -hand side and doing that we would end up with, or to the right -hand side.

04:38 So, y -squared half minus y -t minus 2p squared is equal to 0.

04:50 And now before we solve for y, what i'm going to do is go ahead and multiply this by 2 just so we don't have to work with fractions.

05:00 So multiplying this whole thing by 2 would give us y -square minus 2, yt minus 4 p squared is equal to zero.

05:14 And now let's go ahead and plug this into the quadratic formula.

05:17 So doing that, we'd get y is equal to.

05:21 So it's negative b.

05:23 So it'd be 2...