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Find the eccentricity of the conic $4(2y-x-3)^2 -9(2x+y-1)^2=80$

Solution :

$4(2y-x-3)^2 = 4x^2-16xy+24x+16y^2-48y+36$

and $9(2x+y-1)^2 = 36x^2+36xy-36x+9y^2-18y+9$

$\therefore 4(2y-x-3)^2 -9(2x+y-1)^2 = 7y^2+60x -52xy-32x^2-30y+27 =80$

Can we have other option available so that we will be able to find the solution more quicker way,

Since this is a conic then the given lines let $L_1 =2y-x-3=0 $ and $L_2 = 2x+y-1=0$ are perpendicular to each other .. Can we use this somehow please suggest... thanks..

Sachin
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3 Answers3

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As the Rotation of the axes does not change the eccentricity of a curve

set $2y-x-3=X,2x+y−1 =Y$ so that the given equation becomes $$4X^2-9Y^2=80\iff \frac{X^2}{\frac{80}4}-\frac{Y^2}{\frac{80}9}=1$$

Now we know, $$b^2=a^2(e^2-1)$$ where $a,b\le a$ are the semi-major & semi-minor axes and $e$ is the eccentricity of the hyperbola

lab bhattacharjee
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With a translation of the origin in the center of the conic (which is the point of intersection of the two lines, you get the equation in the form

$$ 4(-X+2Y)^2-9(2X+Y)^2=80 $$

Now the lines are indeed orthogonal: since $1^2+2^2=5$, you can write the equation as $$ 4\cdot5\left(-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\right)^2 -9\cdot5\left(\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y\right)^2=80 $$ The transformation \begin{cases} \xi=-\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\\ \eta=\frac{2}{\sqrt{5}}X+\frac{1}{\sqrt{5}}Y \end{cases} is a rotation, so you get the canonical form $$ 4\xi^2-9\eta^2=16 $$ or $$ \frac{\xi^2}{4}-\frac{\eta^2}{16/9}=1 $$ Thus you have $$a^2=4,\quad b^2=\frac{16}{9}$$

egreg
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However,

With rotation letting Q = 7y^2 -52xy -32x^2 using a rotation matrix have that:

x = cx' + sy' y = -sx' + cy' x^2 = c^2x'^2 +s^2y'^2 + 2csx'y' y^2 = s^2x'^2 +c^2y'^2 - 2csx'y' xy = -csx'^2 + csy'^2 + (c^2 - s^2)(x'y')

Subbing in these and extracting all x'y' components to equate to zero gives:

-14csx'y' -64csx'y' -52(c^2 - s^2)(x'y') = 0 x'y'(52s^2 - 78cs - 52c^2) = 0 2s^2 - 3cs -2c^2 = 0

using quadratic formula for s

s = 3/4c +or- 5/4c = -1/2c or 2c (dividing both by c) gives:

t = -1/2 or 2 hence theta = 26.57 or 63.43

Subbing either into Q since both render xy component zero (subbing 26.57) gives: Q' = 20y'^2 - 45x'^2

Completing the square and re-arranging gives

5((root2y' - 3(root2)/2)^2)/39 - 5((root3x' - 2(root3)/3)^2)/26 = 1

Which is a hyperbola with a^2 = 39/5 and b^2 = 26/5

We know e^2 = 1 - (b^2)/(a^2) = 1/3

Hence I find e = root(1/3)

To lab Bhatt: Barring a mistake on my part it seems that our eccentricities differ where in your assumption of X and Y e = root(65/81). What have I done wrong? Is it the perpendicular bit that allows for X and Y assumption. Please shed some light if you will. I would suggest that maybe rotation doesn't make a difference to eccentricity as you suggest but the assumption does?

rossmckenzie
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