Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$.

Question

Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$.

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The given conic $7x^2+12xy-2y^2-2x+4y-7=0$ is a hyperbola because when i compare it with $ax^2+2hxy+by^2+2gx+2hy+c=0$ and found $h^2-ab>0$,

Then i tried to change $7x^2+12xy-2y^2-2x+4y-7=0$ into the standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ so that length of latus rectum is $\frac{2b^2}{a}$

$7x^2+12xy-2y^2=7x^2-2(-6xy+y^2)=7x^2-2(9x^2-2.3x.y+y^2)+18x^2$

$=25x^2-2(3x-y)^2$

$7x^2+12xy-2y^2-2x+4y-7=25x^2-2(3x-y)^2-2x+4y-7$

Let $x=X,3x-y=Y$ so that $y=3X-Y$

$25x^2-2(3x-y)^2-2x+4y-7=0$ becomes $25X^2-2Y^2-2X+4(3X-Y)-7=0$

$25X^2-2Y^2+10X-4Y-7=0$

$(25X^2+10X+1)-2(Y^2+2Y+1)+2-1-7=0$

$(5X+1)^2-2(Y+1)^2=6$

$\frac{(5X+1)^2}{6}-\frac{(Y+1)^2}{3}=1$

$\frac{(5x+1)^2}{6}-\frac{(3x-y+1)^2}{3}=1$

Now this is in the standard form,So $a^2=6,b^2=3$ and length of latus rectum is $\frac{2b^2}{a}=\frac{6}{\sqrt6}=\sqrt6$

But the answer given is $\sqrt{\frac{48}{5}}$I dont know where i am wrong?

Answer 1

All of your algebraic work is correct, but you will want a different approach. The problem is in your conclusion, when you change variables back to $x,y$:

$$\frac{(5x+1)^2}{6}-\frac{(3x-y+1)^2}{3}=1$$ Now this is in the standard form, so $\ldots$

This is not in standard form because the numerator of the second fraction has both $x$ and $y$ terms. Furthermore, the change of variables will change the foci, coördinates of the latera recta, and so on, and these coördinates do not scale linearly with the variables.

Instead, try a change of variables that consists of a rotation without scaling: $$x = X\cos\theta - Y\sin\theta,\quad y=X\sin\theta + Y\cos\theta$$ where $$\cot2\theta = \frac{A-C}B.$$ You can read more details about this transformation in this excerpt from Stewart. Note that you don't actually need to solve for $\theta$ since we only care about $\cos\theta$ and $\sin\theta$; in your case, you should end up with $$x = 2\sqrt\frac15X - \sqrt\frac15Y, \quad y = \sqrt\frac15X + 2\sqrt\frac15Y.$$ You should be able to take it from here!