Question

let the conics x2=xy+1 and x2=xy-1 have eccentricities e1 and e2, then relation between e1 and e2 is

• A. e1=e2
• B. e1>e2
• C. e1<e2
• D. 1/e1^2+1/e2^2=1

Answer 1

Answer: (D)

1/e1^2+1/e2^2=1

Answer 2

The given equations of the conics are:

1. $x^2 = xy + 1 \implies x^2 - xy - 1 = 0$

2. $x^2 = xy - 1 \implies x^2 - xy + 1 = 0$

These are general second-degree equations of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

For equation 1: $A=1$, $B=-1$, $C=0$, $D=0$, $E=0$, $F=-1$.

For equation 2: $A=1$, $B=-1$, $C=0$, $D=0$, $E=0$, $F=1$.

The nature of the conic is determined by the discriminant $B^2 - 4AC$.

For both equations, $B^2 - 4AC = (-1)^2 - 4(1)(0) = 1 > 0$.

Since $B^2 - 4AC > 0$, both conics are hyperbolas.

To find the eccentricity of a hyperbola $Ax^2 + Bxy + Cy^2 + F = 0$ (when $D=E=0$), we can use the eigenvalues of the matrix $\begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}$. The characteristic equation is $\det \begin{pmatrix} A-\lambda & B/2 \\ B/2 & C-\lambda \end{pmatrix} = 0$, which is $(A-\lambda)(C-\lambda) - (B/2)^2 = 0$, or $\lambda^2 - (A+C)\lambda + AC - B^2/4 = 0$.

The eigenvalues $\lambda_1, \lambda_2$ are related to the coefficients of the transformed equation $A'X^2 + C'Y^2 + F = 0$ (where $A'=\lambda_1$ and $C'=\lambda_2$) after rotating the axes to eliminate the $xy$ term. The equation becomes $\lambda_1 X^2 + \lambda_2 Y^2 + F = 0$. Since it's a hyperbola, $\lambda_1$ and $\lambda_2$ have opposite signs. The standard form of a hyperbola is $\frac{X^2}{a'^2} - \frac{Y^2}{b'^2} = \pm 1$ or $\frac{Y^2}{a'^2} - \frac{X^2}{b'^2} = \pm 1$.

The transformed equation is $\lambda_1 X^2 + \lambda_2 Y^2 = -F$.

The coefficients of $X^2$ and $Y^2$ in the standard form $\frac{X^2}{a^2} \pm \frac{Y^2}{b^2} = 1$ are related to $\lambda_1, \lambda_2, F$.

The equation is $\frac{\lambda_1}{-F} X^2 + \frac{\lambda_2}{-F} Y^2 = 1$.

Let $A'' = \frac{\lambda_1}{-F}$ and $C'' = \frac{\lambda_2}{-F}$. The equation is $A'' X^2 + C'' Y^2 = 1$.

For a hyperbola, $A''$ and $C''$ have opposite signs. If $A">0$ and $C"<0$, the equation is $\frac{X^2}{1/A''} - \frac{Y^2}{-1/C''} = 1$, so $a^2 = 1/A''$ and $b^2 = -1/C''$. The eccentricity is $e^2 = 1 + b^2/a^2 = 1 + \frac{-1/C''}{1/A''} = 1 - A''/C'' = 1 - \frac{\lambda_1/-F}{\lambda_2/-F} = 1 - \lambda_1/\lambda_2$.

If $A''<0$ and $C">0$, the equation is $\frac{Y^2}{1/C''} - \frac{X^2}{-1/A''} = 1$, so $a^2 = 1/C''$ and $b^2 = -1/A''$. The eccentricity is $e^2 = 1 + b^2/a^2 = 1 + \frac{-1/A''}{1/C''} = 1 - C''/A'' = 1 - \frac{\lambda_2/-F}{\lambda_1/-F} = 1 - \lambda_2/\lambda_1$.

In general, for $\lambda_1 X^2 + \lambda_2 Y^2 + F = 0$,

The eigenvalues are the roots of $\lambda^2 - (A+C)\lambda + AC - B^2/4 = 0$.

For both conics, $A=1, B=-1, C=0$.

$\lambda^2 - (1+0)\lambda + (1)(0) - (-1)^2/4 = 0$

$\lambda^2 - \lambda - 1/4 = 0$

$4\lambda^2 - 4\lambda - 1 = 0$.

The eigenvalues are $\lambda = \frac{4 \pm \sqrt{16 - 4(4)(-1)}}{8} = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$.

Let $\lambda_1 = \frac{1+\sqrt{2}}{2}$ and $\lambda_2 = \frac{1-\sqrt{2}}{2}$. Note that $\lambda_1 > 0$ and $\lambda_2 < 0$.

For conic 1: $F=-1$. The transformed equation is $\lambda_1 X^2 + \lambda_2 Y^2 - 1 = 0$, or $\lambda_1 X^2 + \lambda_2 Y^2 = 1$.

$\frac{1+\sqrt{2}}{2} X^2 + \frac{1-\sqrt{2}}{2} Y^2 = 1$.

$\frac{X^2}{\frac{2}{1+\sqrt{2}}} + \frac{Y^2}{\frac{2}{1-\sqrt{2}}} = 1$.

$\frac{X^2}{2(\sqrt{2}-1)} + \frac{Y^2}{-2(\sqrt{2}+1)} = 1$.

$\frac{X^2}{2(\sqrt{2}-1)} - \frac{Y^2}{2(\sqrt{2}+1)} = 1$.

This is a hyperbola of the form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$, with $a^2 = 2(\sqrt{2}-1)$ and $b^2 = 2(\sqrt{2}+1)$.

The eccentricity $e_1$ is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{2(\sqrt{2}+1)}{2(\sqrt{2}-1)} = 1 + \frac{\sqrt{2}+1}{\sqrt{2}-1} = 1 + \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)(\sqrt{2}+1)} = 1 + \frac{2+1+2\sqrt{2}}{2-1} = 1 + 3+2\sqrt{2} = 4+2\sqrt{2}$.

For conic 2: $F=1$. The transformed equation is $\lambda_1 X^2 + \lambda_2 Y^2 + 1 = 0$, or $\lambda_1 X^2 + \lambda_2 Y^2 = -1$.

$\frac{1+\sqrt{2}}{2} X^2 + \frac{1-\sqrt{2}}{2} Y^2 = -1$.

Multiply by -1: $-\frac{1+\sqrt{2}}{2} X^2 - \frac{1-\sqrt{2}}{2} Y^2 = 1$.

$\frac{-(1+\sqrt{2})}{2} X^2 + \frac{-(1-\sqrt{2})}{2} Y^2 = 1$.

$\frac{\sqrt{2}-1}{2} Y^2 - \frac{\sqrt{2}+1}{2} X^2 = 1$.

$\frac{Y^2}{\frac{2}{\sqrt{2}-1}} - \frac{X^2}{\frac{2}{\sqrt{2}+1}} = 1$.

$\frac{Y^2}{2(\sqrt{2}+1)} - \frac{X^2}{2(\sqrt{2}-1)} = 1$.

This is a hyperbola of the form $\frac{Y^2}{a^2} - \frac{X^2}{b^2} = 1$, with $a^2 = 2(\sqrt{2}+1)$ and $b^2 = 2(\sqrt{2}-1)$.

The eccentricity $e_2$ is given by $e_2^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{2(\sqrt{2}-1)}{2(\sqrt{2}+1)} = 1 + \frac{\sqrt{2}-1}{\sqrt{2}+1} = 1 + \frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)} = 1 + \frac{2+1-2\sqrt{2}}{2-1} = 1 + 3-2\sqrt{2} = 4-2\sqrt{2}$.

We have $e_1^2 = 4 + 2\sqrt{2}$ and $e_2^2 = 4 - 2\sqrt{2}$.

Consider $e_1^{-2} + e_2^{-2} = \frac{1}{4+2\sqrt{2}} + \frac{1}{4-2\sqrt{2}} = \frac{4-2\sqrt{2} + 4+2\sqrt{2}}{(4+2\sqrt{2})(4-2\sqrt{2})} = \frac{8}{16 - (2\sqrt{2})^2} = \frac{8}{16 - 8} = \frac{8}{8} = 1$.

So, $e_1^{-2} + e_2^{-2} = 1$, which means $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.