# Thread: Tangent to a curve

1. ## More help with differentiation please!

1. The equations of a tangent and normal to the curve xy=4 at the point where x=2 are y=4-x and y=x.
Show that the normal does intersect the curve again and find the coordinates of the point of intersection.

2. Find the equations of the normals to the curve xy=3 which are parallel to the line 3x-y-2=0

Thanks so much if you can help!

2. " 1. The equations of a tangent and normal to the curve xy=4 at the point where x=2 are y=4-x and y=x.
Show that the normal does intersect the curve again and find the coordinates of the point of intersection."

If two curves intersect, then their coordinates at their point of intersection are the same.
xy = 4 ----(1)
y = 2 ------(2)

y of (1) = y of (2),
4/x = x
4 = x^2
x = +,-sqrt(4) = +,-2
That means the two curves intersect at x=2 and x = -2.
At x = -2,
y = x = -2
So, (-2,-2) is the other point of intersection.

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2. Find the equations of the normals to the curve xy=3 which are parallel to the line 3x-y-2=0

Slope of tangent lines to xy=3:
Differentiate both sides with respect to x,
x(dy/dx) +y(1) = 0
dy/dx = -y/x --------slope of tangent lines to xy=3 curve.

3x -y -2 = 0
-y = -3x +2
y = 3x -2 ------y=mx+b form.
So, slope is 3.
Any line parallel to [3x -y -2 = 0] will have a slope of 3 also.
So if the normals to the xy=3 curve are parallel to [3x -y -2 = 0], then their slope is 3 also.

The normals and tangent lines to xy=3, at the same points, have slopes that are negative reciprocals.
Meaning, if the slope of the normal line is 3, then the slope of the corresponding tangent line is -1/3.
So,
-y/x = -1/3
Cross multiply,
-y*3 = x*(-1)
-3y = -x
3y = x
y = x/3 -----------the normal lines to xy=3 curve.

Doing it in another way:

If the slope of the tangent lines to xy=3 is -y/x, then the slope of the normal lines is -1 / (-y/x) = x/y

x/y = 3
x = 3y
y = x/3 -------normal lines to xy=3.

3. Thanks so much. You're method for the last question seemed really logical however the answer is supposed to be y=3x+/-8. Do you know why this may be?

4. " You're method for the last question seemed really logical however the answer is supposed to be y=3x+/-8. Do you know why this may be? "

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Umm, that made me review my answer yesterday. And, yes, the correct answer is y = 3x +8 , or y = 3x -8.

In my answer yesterday I stopped at y = x/3.
What I did not figure out then is that at the point of normalcy (?) y = x/3 only. The equation of the normal line is not yet the y = x/3.

So, let me finish the answer yesterday.

At the point of "normalcy", y = x/3, so, substitute that into xy=3,
x(x/3) = 3
x*x = 3*3
x^2 = 9
x = +,-sqrt(9)
x = +,-3 -------the x's at the points of normalcy.

The y's:

When x = 3,
xy = 3
(3)y = 3
y = 3/3 = 1
So, (3,1) is one point of normalcy.

When x = -3,
xy = 3
(-3)y = 3
y = 3/(-3) = -1
So, (-3,-1) is the other point of normalcy.

For the equations of the two normal lines:

At (3,1), and with a slope of 3, by point-slope form,
y-1 = 3(x-3) ----------answer.
Or,
y -1 = 3x -9
y = 3x -9 +1
y = 3x -8 -----------in y=mx+b form, answer.

At (-3, -1), and with a slope of 3, by point-slope form,
y -(-1) = 3(x -(-3))
y +1 = 3(x+3) ----------answer.
Or,
y +1 = 3x +9
y = 3x +9 -1
y = 3x +8 -----------in y=mx+b form, answer.