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# Prove that the curve $2x^{2}+4y^{2}=1$ and $6x^{2}-12y^{2}=1$ cut each other at right angles.

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• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) • If the curves y=f(x),y=g(x) intersect at (x_1,y_1) then the angle of intersection of the curves is the angle between the tangents at the point (x_1,y_1) • If m_1=t'(x_1) and m_2=g'(x_1) then the angle is \tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2} when the curves are orthogonal , the tangents are perpendicular to each other and m_1m_2=-1 Step 1: Let the curves 2x^2+4y^2=1---(i) and 6x^2-12y^2=1---(ii) intersect at (x_1,y_1) 2{x_1}^2+4{y_1}^2=1 6{x_1}^2-12{y_1}^2=1 Solving for ({x_1}^2,{y_1}^2) \Delta =\begin {vmatrix} 2 & 4 \\ 6 & -12 \end {vmatrix}=-24-24=-48 \Delta_1 =\begin {vmatrix} 1 & 4 \\ 1 & -12 \end {vmatrix}=-12-4=-16 \Delta_2 =\begin {vmatrix} 2 & 1 \\ 6 & 1 \end {vmatrix}=2-6=-4 {x_1}^2=\large\frac{\Delta _1}{\Delta }=\frac{1}{3}$$=>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.