Question

find the angle of intersection between the curve x^2-y^2=a^2 and x^2+y^2=a^2 (sqr rtof 2)?pls help me

Answer 1

\[} find the \angle of intersection \between the curve x^{2}-y^{2}=a^{2} and x^{2}+y^{2=a^{2}\]

Answer 2

solution: i got slope m1= dy/dx=x/y and slope m2=dy/dx= (-x/y) and \[\Theta=\tan^{-1}(2xy/y^{2}-x^{2})\]

Answer 3

the points of intersection of the curves are at the \[\pm \sqrt{a}\]

Answer 4

but how do i get ans: \[ \Theta=\tan^{-1}(\pm a^2/-a^{2})\]

Answer 5

i got the dinomtr as -a^{2}

Answer 6

so are you asking about the tangents to the curves at these points...
as they are vertical lines

\[x = \pm a\]

Answer 7

yep

Answer 8

how do i get the numerator 2xy = pm a^{2}

Answer 9

oops just reread

Answer 10

from the question we have to fine the theta

Answer 11

i got theta = tan^{-1} (2xy/y^{2}-x^{2})

Answer 12

\[\theta = \tan^{-1} (2xy/y^{2}-x^{2}) = \tan^{-1}(\pm a^{2}/-a^{2}\]

Answer 13

i gotthe dinomntr -a^{2} from the question'\[\X^{2}-y^{2}\]\

Answer 14

pls hlp me

Answer 15

how do i get the numeratr \[\ pm a^{2}\]\

Answer 16

\[\x^{2}+Y^{2}=a^{2}sqrt{2}\]\

Answer 17

\[\(x+y)^{2}-2xy= a^{2}sqrt{2}\]\

Answer 18

how to solve aftr ths to get 2xy = \[\pm a^{2}\]\

Answer 19

are these your equations?
x^2-y^2=a^2 and x^2+y^2=a^2

Answer 20

no the question s

Answer 21

i don't understand that last line in your question.

Answer 22

x^2-y^2 =a^2 and x^2+y^2=a^2 sqrt{2}

Answer 23

I think you need a nominal value for a... perhaps a = 1

Answer 24

first one is a hyperbola, second one is a circle with slightly bigger radius than the transverse and conjugate axis of the hyperbola. they will intersect at 4 points. and you're looking for the angle in which the tangent lines intersect at these points?

Answer 25

yes sir

Answer 26

because of symmetry, i don't think that would be too hard eh?
add those two equations to get rid of the y^2.... what do you get for x?

Answer 27

this s an example sum in my book it s given theta= tan^{-1} (2xy/y^2-x^2)=tan^-1(pm a^2/-a^2)

Answer 28

is x =a sqrt (1+ sqrt2)/sqrt2

Answer 29

sorry man, that thing you just put up don't make sense to me.
i was gonna do it with finding the slope of the tangent lines at those intersections and using those slopes to calculate the angle between them.

Answer 30

yes it s to find the angle btween them

Answer 31

i got the slopes m1=x/y and m2=(-x/y)

Answer 32

by substituting in formula theta =tan^{-1} [(m1-m2)/(1-m1m2)] i got tan^{-1}(2xy/y^2-x^2)

Answer 33

sorry formula tan^{-1}[(m1-m2)/1+m1m2]

Answer 34

i got a different x-coordinate for my intercept