Step 1:

Let the curves $2x^2+4y^2=1$---(i) and $6x^2-12y^2=1$---(ii) intersect at $(x_1,y_1)$

$2{x_1}^2+4{y_1}^2=1$

$6{x_1}^2-12{y_1}^2=1$

Solving for $({x_1}^2,{y_1}^2)$

$\Delta =\begin {vmatrix} 2 & 4 \\ 6 & -12 \end {vmatrix}=-24-24=-48$

$\Delta_1 =\begin {vmatrix} 1 & 4 \\ 1 & -12 \end {vmatrix}=-12-4=-16$

$\Delta_2 =\begin {vmatrix} 2 & 1 \\ 6 & 1 \end {vmatrix}=2-6=-4$

${x_1}^2=\large\frac{\Delta _1}{\Delta }=\frac{1}{3}$$=>x_1=\pm \large\frac{1}{\sqrt 3}$

${y_1}^2=\large\frac{\Delta _2}{\Delta }=\frac{4}{48}=>y_1=\pm \frac{1}{2 \sqrt 3}$

The four points of intersection are given by $\bigg(\pm \large\frac{1}{\sqrt 3}$$,\pm \large\frac{1}{2 \sqrt 3 }\bigg)$

Step 2:

We consider the point of intersection $\bigg(\large\frac{1}{\sqrt 3},\frac{1}{2 \sqrt 3 }\bigg)$

We need to find the slope of the tangents to the curves at this point.

Differentiating (i) and (ii) with respect to x

$4x+8y \large\frac{dy}{dx}$$=0=>\large\frac{dy}{dx}=\frac{-x}{2y}$

$12x-24y \large\frac{dy}{dx}$$=0=>\large\frac{dy}{dx}=\frac{x}{2y}$

At $\bigg(\large\frac{1}{\sqrt 3},\frac{1}{2 \sqrt 3 }\bigg)$ the slope of the tangent to the two curves are

$m_1=\large\frac{\Large\frac{-1}{\sqrt 3}}{\Large\frac{2}{2 \sqrt 3}}$$=-1$

$m_2=\large\frac{\Large\frac{1}{\sqrt 3}}{\Large\frac{2}{2 \sqrt 3}}$$=1$

Now $m_1m_2=-1$

The two tangent intersect at right angles.

Therefore the curves cut each other at right angles.

By symmetry the curves intersect at right angles at the other three points is also.