00:02
We have three parts in this problem.
00:04
First part, we want to find an equation of the tangent line and an equation of the normal line to the circle x square plus y square equal 169 at the point negative 512.
00:20
Part 2, we want to find an equation of the tangent line and an equation of the normal line to the curve defined implicitly by the equation x square plus 10 xy plus 4 y square plus 5 equals 0 at the point 1 added 1 and in the third part we do a similar exercise to the second one for the curve defined implicitly implicitly by the equation y square equal 5 4 5 x to the 4th minus x square at the point 1 negative 2.
00:55
So let's start with part 1.
01:00
And in part 1, you have a circle center at 0 -0 because the equation is x -square plus y -square equal 169.
01:10
In fact, 169 is 13 square.
01:14
So the center of the circle is 0 -0, and the radius is 13.
01:22
So you have something like this.
01:26
More or less and you have here 13 okay 13 and negative 13 and 13 and 13 and 19 13 here that's it so at any point on the circle on the circumference of the circle the tangent line at that point will be perpendicular to the radius at that point right the design here is not well done but that's the case that is if we draw the line joining the center of the circle that is zero zero and a point on the circle on the circumference of the circle that line that segment is perpendicular to the tangent line let's see if i draw it here better something like this and that happens at any point on the circumference of the circle so it's easy to find the tangent line equation by simply calculating first the slope of the line passing through the center of the circle and the point of tangency on the circumference of the circle.
03:00
I mean that.
03:02
The equation in this case, well, a thing we got to verify first is very important.
03:10
Is at the point we are considering in part one, that is this point negative 512, is a point on the circumference of the circle.
03:21
That is, we get to verify that this point satisfies the equation of the circle.
03:26
And that's true because negative 5 square plus 12 square is 25 plus 144, and that is 169.
03:39
So it's very important to verify that.
03:41
Because we must calculate all the equations on points that are just on the circle or on the circumference of the circle rather so that's it we know that point is on the circle and we can calculate the normal and the tangent lines so the line the line that say here passes through the center of the circle of the circle passes through let me write this again it passes through the center of the circle 0 0 and the given point on the circle negative 512 okay so if we know that that line passes through these two points we can calculate it's a slope.
05:16
The slope of the normal line is m equal 12 minus 0 over negative 5 minus 0.
05:43
12 minus 0 over negative 5 minus 0 and that is negative 12 over 5 over 5.
05:53
And because the normal line passes through 0 0, so the normal line equation is y minus 0 equals slope negative 12 over 5 times x minus 0 because it passes through 0.
06:22
And that is y equal negative 12 over 5 times x.
06:30
That's the equation of normal line.
06:34
So let's put here normal line at negative 5 at the 5 12 and because the tangent line is perpendicular to that normal tangent line at negative 512 is perpendicular to the normal.
07:38
We have that the slope of this tangent line at negative 512 is let's call it m is negative 1 over the slope of the normal that is negative 12 over 5 5 that is negative 12 over 5 that is because the slope or said another way the slope of perpendicular lines multiply by one to each other is negative one.
08:37
There is a multiplication of the slopes of perpendicular lines in negative one.
08:42
So one of them can be calculated as the reciprocal of the other change of sign or change in signs.
08:51
So we get negative one over the slope of the normal and that will give us five over 12.
08:59
That's the slope of the tangent line to the circle at negative 512.
09:06
And so because it passes through since the tangent line passes through the point of tangency, it's 512, you can write the equation of that land.
09:24
And the line to the circle at negative 512 passes through this point.
09:43
That is negative 512 we have that equation of the tangent line to the circle at negative 512 is y minus 12 equal slope which we calculated here 5 of 12 times x minus 5 that is y minus 12 equal slope which we calculated here 5 over 12 times times x minus negative 5 that is minus 12 equal 5 over 12 times x plus 5 and so y is sorry so by 12 both sides we get 12 y minus 12 times 12 is 144 equal 5x plus 25 and so 5x minus 12 y minus 12 y and 144 plus 25 is 169 minus plus 169 equals 0.
11:36
So that's the equation of the tangent line to the circle at the point negative 512.
11:49
And this slope can be found, in fact, using implicit differentiation in the equation of the circle.
11:59
Let's verify that.
12:04
And that's the idea we will use in the other curves in which we cannot use the geometrical fact we use here.
12:13
Remember, we said here that geometrically we know, because this is a circle, that the line passing through the center of the circle and the point on the circle is always a normal to the circle at that point.
12:31
Is perpendicular to the tangent line.
12:35
So we use geometry to imply that here, but you can do implicit differentiation in the equation of the circle given here, and you find the derivative of the both sides of the equation respect to x.
13:03
So the derivative respect to x of the left -hand side of the equation has to be equal to the derivative respect to the x after the right hand side of the equation and the derivative of x square plus square is the derivative of x square respect to x is 2x but here when we find the derivative of y square we got to be careful that y depends on x so we get 2 y times the derivative of y respect to x here we have applied the chain rule and that's equal to zero because 169 is a constant okay so so here we can find the derivative of y respect to x is negative the 2x over 2y.
13:57
Can divide by y because y can be 0 or if y is 0 we cannot use this simplest differentiation in any case.
14:08
So we get negative x over y and we have this.
14:15
So the derivative of y respect to x is negative x over y.
14:20
Each time y is not zero.
14:23
If y is zero, we must use another fact to find derivative.
14:29
Okay, so this means that that derivative add the point xy equal negative 512, which is a point given on circle in this problem in part one, is negative x that is negative 5 over 12.
14:48
Negative negative 5 is 5 over 12 and that's just the slope remember the derivative at the point is the slope of the 10 line at that point so we found just what we found here using geometry but that way of finding the slope of the tangent line will be the way we get to use in the next two parts because we don't have geometry involved or at least not evidently using the given equation so in part two we do that we have the implicit definition of the curve we want to find again equations for the normal on the tangent lines the curve is given implicitly but this equation and the point is one negative one you get a verified first and one negative one is on that curve and that's true because one square is one four times y square negative one square is one so we get one plus four plus five that is one here four here and five here is ten but this term will be negative ten because it's ten times x plus y and y is negative one so we get negative ten so the sum of all the terms is zero so it's true that is the point one negative one is on the curve because it satisfies the equation defining implicitly the curve.
16:30
And so now i'm going to do implicit differentiation to the equation of the curve.
16:38
Let's write it again to have it here.
16:40
X square plus 210 xy plus 10.
16:49
Sorry 10 x y plus 4 y square plus 5 that's it that's it equals 0 so we do implicit differentiation respect to x so the derivative respect to x of the left -hand side of the equation x squared plus 10 xy plus 4 y square plus 5 is derivative respect to x of 0 and we know 1 y depends on x so we get to apply the chain rule each time we find y on the formula.
17:40
And then we have 2x plus 10 times and now here we have a product of two functions depending on x, that is x itself and y which depends on x.
17:55
So we have a product and we applied the derivative of product.
18:00
So we get derivative of x respect to x times y...