The eccentricity of the conic 3 x 2 + 3 y 2 − 2 x y − 2 = 0 is e , find e 2 .
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A Solution without rotation of axis .
Partially Differentiating The Equation of ellipse once wrt x and then wrt y we obtain centre of ellipse as origin. (It can be checked by using condition for an equation to represent an ellipse)
Ellipse Being a central curve every chord passing through its centre is bisected at the centre of ellipse.
We Assume a chord of ellipse passing through origin as y = mx and m = tan(t) wherein t is the angle made by chord from positive x axis in anticlockwise direction.
Using Parametric Coordinates Of Line any point on the line at a distance r from origin can be taken as (rsint,rcost).
Now For This chord to be the major or minor axis of ellipse the parameter r should be maximum or minimum accordingly.
Substituting this point into equation of ellipse we obtain r^2 = 2 / 3-sin(2t)
Now we can easily Maximise and Minimise r which will give the value of semi major and minor axis of ellipse.
Hence We will get the eccentricity of ellipse
for a better approach observe curve is symmetric about y=x.Hence we can easily find lengths of major or minor axes as the other axis will be y=-x. once we get them, we can easily get eccentricity. Note: Use h^2-ab<0 and Δ≠0 to check for ellipse.
This Is Brilliant ! Upvoted :)
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you're in 12th?
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nope in 11th (in two days will come to 12th)
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@Prakhar Bindal – are u in fiitjee ?
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@Ayush Maurya – yeah, i'm in fiitjee jaipur.
@Ayush Maurya – Yeah in fiitjee east delhi!
easy question of rotation of axis.
then post a solution
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The given equation is :- 3 x 2 + 3 y 2 − 2 x y − 2 = 0 Now we'll transform the coordinates by rotating the axis by some θ so that it no longer contains the term x y . Now we'll apply the following transformation:- x = X cos θ − Y sin θ y = X sin θ + Y cos θ So our equation becomes (after simplification) :- X 2 ( 3 − 2 sin θ cos θ ) + Y 2 ( 3 + 2 sin θ cos θ ) + 2 X Y ( sin 2 θ − cos 2 θ ) − 2 = 0 If θ = 4 π , the coefficient of X Y will become 0 . Now our equation becomes:- 2 X 2 + 4 Y 2 = 2 X 2 + 2 Y 2 = 1 This is the same curve as given in the question but only the coordinate system is rotated. So it has the eccentricity as the original conic. This transformed equation has eccentricity :- e = 1 − 2 1 e 2 = 1 − 2 1 e 2 = 2 1