Question

Hi all! Q:"Where does the normal line to x^2 -xy+y^2=3 at P(-1,1) intersect the ellipse a second time?"

Answer 1

I found the normal line to be y=x+2 but I'm not sure what to do next

Answer 2

that equation for the normal line does not sound right.

Did you use implicit differentiation to find dy/dx ?

Answer 3

yes

Answer 4

so I broke it up

Answer 5

chain rule for x^2 and y^2

Answer 6

Turn into 2x and 2yy'

Answer 7

product rule for xy

Answer 8

d/dy (x^2 -xy+y^2=3 )

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

d/dy (x y)= x dy/dx + y

Did you get this

2x - x y' - y + 2y y' = 0

Answer 9

sec

Answer 10

I got a different middle result I'll type it

Answer 11

so product rule of xy is d/dx x(y)-d/dx y(x)

Answer 12

which is then 1*y-y'x

Answer 13

but I have to apply the minus sign

Answer 14

except you do not subtract. d(xy)= x dy + y dx

Answer 15

ah so for product rule you add?

Answer 16

ok gotcha

Answer 17

see https://en.wikipedia.org/wiki/Product_rule

Answer 18

so then the derivative at the point is 1 not -1

Answer 19

you should get
2x - x y' - y + 2y y' = 0

solve for y'
what do you get ?

Answer 20

(-y-2x)/(2y+x)

Answer 21

=y'

Answer 22

?

Answer 23

wait

Answer 24

nvm need to rewrite this section, sec pls :)

Answer 25

so, y'= (-y-2x)/(2y-x)

Answer 26

is this correct?

Answer 27

closer.
\[ 2x - x y' - y + 2y y' = 0 \\
2y y' - x y' = y-2x\\
(2y -x)y'= (y-2x)\\
y'= \frac{y-2x}{2y -x}
\]

Answer 28

now evaluate at P(-1,1)

Answer 29

ok

Answer 30

3/3 =1

Answer 31

so the slope of the normal line is -1/1 =-1 then?

Answer 32

yes. now use point- slope formula

y - y0= m( x - x0) where (x0,y0) is (-1,1)

Answer 33

y-1=-(x+1)

Answer 34

or y=-x

Answer 35

simplify to y = -x

now you want to find where this line intersects the ellipse.
sub in (-x) for y into
x^2 -xy+y^2=3
and solve for the two x values

Answer 36

aha I see

Answer 37

one of the x values will be -1 (because -1,1 intersects the curve)

Answer 38

then it should be 1 right?

Answer 39

once you find the other x value, y = -x (or use the ellipse curve, to find y.. but that is harder)

Answer 40

ok

Answer 41

with y= -x in
x^2 -xy+y^2=3
you get

x^2 + x^2 + x^2 = 3
3x^2 = 3
x^2 = 1
x= ±1
so x=1 is the other x value.

Answer 42

yep just finished, so the point is 1,-1

Answer 43

Here is a graph of your problem

Answer 44

ah ok, cool, nice to visualize. Thanks a lot for your help! :)