Question

Please help me check my work. I needed the equation for the tangent line to the elipse 2x^2 + y^2 = 3 and that was parallel to 2x + y = 5.

Answer 1

Wait a sec.

Answer 2

I first wrote it like this:

r: (x,y) = (x0,y0) + lambda v, where v gradient f(x0,y0) = 0.

gradient of f(x,y) = (4x,2y)
gradient of f(x0,y0) = (4x0,2y0)

Answer 3

Then I did the same with 2x + y = 5

r2: (x,y) = (x1,y1) + t u

gradient of f(x,y) = (2,1)
gradient of f(x1,y1) = (2,1)

u = (-1,2)

v is parallel to (2,1), so for some value of h, v = h (-1,2)

v = (2y0,-4x0)

2y0 = - h
-4x0 = 2h

y0 = -h/2
x0 = -h/2

h = -2

y0 = 1
x0 = 1

Answer 4

r: (x,y) = (1,1) + lambda (2,-4)

Answer 5

differentiate the elipse :

4x + 2y(dy/dx) = 0
dy/dx = -2x/y

the line given is y = -2x + 5

so we need
-2x/y = -2
->
x = y

if we put x = y in the equation we get
3x^2 = 3
x = +-1
and so y = +-1

so there are two options
x=1 , y =1 slope -2
y = 1 -2(x-1)
y=1-2x+2
y = -2x +3

x=-1,y=-1 slope -2
y = -1 -2(x+1)
y = -1 -2 -2x
y = -2x -3

Answer 6

ok?

Answer 7

Is it right?

Answer 8

there are two tangent lines possible as i wrote :
y = -2x +3
y = -2x -3

Answer 9

Sorry, your other post didn't show

Answer 10

oh so you see it now ?

Answer 11

Yes. But is what I did right too? And if it's wrong, why?

Answer 12

you got those two lines ?

Answer 13

I got r: (x,y) = (1,1) + lambda (2,-4)

Answer 14

im not sure that its the same

Answer 15

Let me check: (x,y) = (1 + 2lambda, 1 - 4 lambda)

x = 1 + 2 lambda
y = 1 - 4 lambda

y = -2x + b = - 2 - 4 lambda + 3 = 1 - 4 lambda

b = 3

y = -2x + 3

Answer 16

I don't get the other line with -3.

Answer 17

?

Answer 18

why you got only x,y = 1,1
i got -1,-1 as well

Answer 19

-1,-1? I'll see about it.

Answer 20

y0 = -h/2
x0 = -h/2

h = 2

y0 = -1
x0 = - 1

You're right. But then are there infinite lines? I can choose any h.

Answer 21

Ha! I get it, any other h will be just a multiple of 1 or -1. You're right. Thank you.

Answer 22

i dont really fully understand what you did there
but if it works for you its fine
you see we could do this without using any vector analysis

Answer 23

You see, any other h, will just be a constant times 1 or - 1. The first curve is an ellipse, and therefore it will have to tangents at two simetrical points parallel to 2x + y = 5.

Answer 24

ok but i would like you to explain (if you want) your method

i understand you took the gradient of the line 2x + y = 5
its (2,1)

i dont know why then you chose u(-1,2)

Answer 25

Because the dot product of the gradient times u = 0.

Answer 26

but then you find a perpendicular line ?

Answer 27

The gradient is perpendicular to the vector that gives the direction of the tangent line.

Answer 28

isnt the gradient tangent to the curve ? :|

Answer 29

No, it's perpendicular to it. It's only tangent if you have the curve of maximum angle.

Answer 30

well im confused lol i have to think about it .. thank you

Answer 31

You're welcome, and thank you for helping me out.