Ellipse, hyperbola and principle axis 
Would anyone mind telling me how to solve (a)? I have no idea what I should do to solve this problem.
Also, what is principal axes?
 A: I think $x_1$ and $x_2$ are the co-ordinate axis . I am going by the following approach , which is a little long but this method works . 
Seeing the $x_1x_2$ term I figured out that the principal axis of this curve are not perpendicular to the co-ordinate axes . Hence I supposed the equation of the curve of the form $$a(x+ty)^2+b(tx-y)^2=1$$
Now I have taken that the principal axes pass through the origin because there are no $x_1$ and $x_2$ term of single power . And the principal axes should be perpendicular to each other hence I have taken the slope of one to be $t$ and other to be $-\frac{1}{t}$ (the product of slopes of two perpendicular lines is equal to $-1$) 
Now we are all set to go , expanding the above equation we will get ( work out the steps , I am writing the final result ) 
$$(a+bt^2)x_{1}^2+2t(a-b)x_1x_2+(at^2+b)x_{2}^2 = 1$$
Now reducing the original equation to this form you will get 
$$ \frac{6}{14}x_1^2-\frac{4}{14}x_1x_2+\frac{3}{14}x_2^2 =1 $$
Equating the co-efficients you will get 
$$a+bt^2 = \frac{6}{14}\ \ \ \ \ \ \ \ \ \ \ \ \ \cdots(1)$$
$$2t(a-b)=-\frac{4}{14}\ \ \ \ \ \ \cdots(2)$$
$$at^2+b=\frac{3}{14}\ \ \ \ \ \ \ \ \ \ \ \ \ \cdots(3)$$
Now subtract $(3)$ from $(1)$ you will get an equation in $(a-b)$ and $t$ . Using the $(2)$ equation put the value $(a-b)$ in terms of $t$ you will get an equation in $t$ , then use that value of $t$ in $(1)$ and $(2)$ to get $a$ and $b$ . 
This way you get the value of $t$ hence the equation of principal axes and value of $a$ and $b$ come to be positive , hence it is an ellipse . 
