ellipse
i do have a soln for this but can anyone explain how are parametric points assumed
29 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.what the hell
rcosθ rsinθ is for circle
this gives correct answer tho
what how
it says π/sqrt(5)
but value of A
u get 3
which is correct
weird
can't get anything
Integrate karne ka socha par y as a function of x bhi nahi aaraha sahi se
This is just piab.
The area of ellipse.
a and b nahi nikal rahe
😿
Got it
Again, handwriting is bad. Please forgive.
@hardcoreisdead @Dexter
wtf is this manipulation
nah
😭😭😭
nice manipulation
but i wanna know why in my method parametric point is rcostheta, rsintheta since parametric point in ellipse is acostheta,bsintheta
x+y=sqrt(2)/sqrt(5)cos(theta) and x-y=sqrt(2)sin(theta) from this. And solving both the equations give you x=csin(theta) and y=ccos(theta) respectively.
It only worked because the axis in the question were x+y and x-y, which give such result ig.
oh yeah that could make sense
if i were to get another ellipse where x=y and x=-y arent major and minor how do i use this method
or should i just leave that for now since all of this will be in area under curve
It won't. It's a pain to integrate these
And God forbid anything about perimeter of an ellipse
Somebody challenged me to do it and i discovered that's it's actually impossible
Well, analytically at least
You're better off simplifying the eqn.
What I've done here basically
1. Find Center
2. Replace x by x-h and y by y-k
3. You have two perpendicular lines. Consider them as (y+bx) and (by-x). Now try to resolve.
thank you so much
2x^2 + 6xy+5y^2=1
how can we get area of this using the approach above ??
same doubt...in the original ques, x2 and y2 had same coeff so it was easy
@SirLancelotDuLac ..
As far as I could firgure out, One method is to assume the axes to be ax+y and ay-x and then assume equation and try to find out the answer.
Another method is writing it as $(x^{2}+4xy+y^{2})+(x^{2}+y^{2}+2xy)=1$ implying $\frac{(\frac{x+2y}{\sqrt{5}})^{2}}{\frac{1}{5}}+\frac{(\frac{(x+y)^{2}}{\sqrt{2}})}{\frac{1}{2}}=1$
SirLancelotDuLac
Now note in the second one axes are not perpendicular to each other. So you will have to find the ratio in areas of quadrilateral (0,0),(0,1),(1,1) and (1,0) in conventional axes and this new axes one. This is the ratio of piab and the new area
And that is the value of 2x2 determinant of the coeffecient of equations.
In any case pretty sure you won't get such question in any test (where factoring the elipse equation into perpendicular axes is that hard).
not even adv?
how are axes not perpendicular of en ellipse
You can apply transformation for a graph (for area) as long as axes are not collinear.
we good?
+solved @SirLancelotDuLac
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