Question

Find the equation to the common conjugate diameters of the conics (1) $x^2+4xy+6y^2=1$ and $2x^2+6xy+9y^2=1$, and (2) $2x^2-5xy+3y^2=1$ and $2x^2+3xy-9y^2=1$.

Answer 1

To find the equation to the common conjugate diameters of the conics

1) The equations of the conic are

$x^2+4xy+6y^2=1$ .... $\left(1\right)$ and $2x^2+6xy+9y^2=1$ ... $\left(2\right)$

We know that, if the straight lines $A{x}^2+2Hxy+B{y}^2$ are the conjugate diameters of $a{x}^2+2hxy+b{y}^2$ then, $aB-2hH+bA=0$

Now, let the straight lines $A{x}^2+2Hxy+B{y}^2=0$ ..... $\left(3\right)$ be the conjugate diameters of both $\left(1\right)$ and $\left(2\right)$

$\therefore 6A-4H+B=0$ ........ $\left(4\right)$

and $9A-6H+2B=0$ ...... $\left(5\right)$

Solving $\left(4\right)$ and $\left(5\right)$, we get

$\frac{A}{-8+6}=\frac{H}{9-12}=\frac{B}{-36+36}$

$\frac{A}{-2}=\frac{H}{-3}=\frac{B}{0}=k$ ( Say )

$\Rightarrow A=-2k,H=-3k,B=0$

Putting these values in $\left(3\right)$, we have

$-2k{x}^2-6kxy=0$ or $k(x+3y)=0$

Hence, $(x+3y)=0$

2) The equations of the conic are

$2x^2-5xy+3y^2=1$ .... $\left(6\right)$ and $2x^2+3xy-9y^2=1$ ... $\left(7\right)$

Now, let the straight lines $A{x}^2+2Hxy+B{y}^2=0$ be the conjugate diameters of both $\left(6\right)$ and $\left(7\right)$

$\therefore 3A+5H+2B=0$ ........ $\left(8\right)$

and $-9A-3H+2B=0$ ...... $\left(9\right)$

Solving $\left(8\right)$ and $\left(9\right)$, we get

$\frac{A}{10+6}=\frac{H}{-18-6}=\frac{B}{-45+9}$

$\frac{A}{16}=\frac{H}{-24}=\frac{B}{-36}=k$ ( Say )

$\Rightarrow A=16k,H=-24k,B=-36k$

Putting these values in $\left(3\right)$, we have

$16k{x}^2-48kxy-36k{y}^2=0$ or $4x^2-12xy-9y^2=0$