If the curves $ {{y}^{2}}=4ax\text{ and xy=}{{\text{c}}^{\text{2}}}, $ cut orthogonally then $ {{c}^{2}}=32{{a}^{4}} $

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Hint: Assume any arbitrary point of intersection of parabola $ {{y}^{2}}=4ax $ and hyperbola $ \text{xy=}{{\text{c}}^{\text{2}}} $ as $ P({{x}_{1}},{{y}_{1}}) $ Find the slopes of both curves at point $ P({{x}_{1}},{{y}_{1}}) $ as $ {{m}_{1}}\text{ and }{{\text{m}}_{\text{2}}} $ then with the help of orthogonality condition $ {{m}_{1}}\times {{\text{m}}_{\text{2}}}=-1 $ give the proof.

Complete step-by-step answer:
We have a parabola $ {{y}^{2}}=4ax $ and a hyperbola $ \text{xy=}{{\text{c}}^{\text{2}}} $ cutting each other orthogonally or perpendicularly.

  & {{y}^{2}}=4ax \\
 & \text{Focus at (a,o)} \\

\[\text{Hyperbola: yx=}{{\text{c}}^{\text{2}}}\]
Intersection of above two curves:

Suppose, they intersect at $ P({{x}_{1}},{{y}_{1}}) $ orthogonally.
Then, the tangent at point $ P({{x}_{1}},{{y}_{1}}) $ to both the curves should be $ {{\bot }^{r}} $ (perpendicular) to each other.
\[\Rightarrow {{T}_{1}}{{\bot }^{r}}{{T}_{2}}\]
Let’s suppose the shape of tangent line $ {{T}_{1}} $ is $ {{m}_{1}} $ and slope of tangent line $ {{T}_{2}} $ is $ {{m}_{2}} $ at point $ P({{x}_{1}},{{y}_{1}}) $
According to orthogonal condition:
\[\Rightarrow {{m}_{1}}\times {{\text{m}}_{\text{2}}}=-1\]
Thus, we find the slope of both the tangents through their curve.
 $ {{T}_{1}} $ corresponds to curve $ {{y}^{2}}=4ax $
Slope $ {{m}_{1}} $ is nothing but $ \dfrac{dy}{dx} $ therefore, differentiating $ {{y}^{2}}=4ax $ with respect of x.
  & \Rightarrow 2y\dfrac{dy}{dx}=4a \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}\Rightarrow {{m}_{1}}=\dfrac{2a}{y} \\
\[{{m}_{1}}\text{ at point }P({{x}_{1}},{{y}_{1}})\Rightarrow {{m}_{1}}=\dfrac{2a}{{{y}_{1}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (1)}\]
Now, $ {{T}_{2}} $ corresponds to curve $ \text{xy=}{{\text{c}}^{\text{2}}} $
\[{{m}_{2}}\Rightarrow \dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( {{c}^{2}} \right)\]
Applying chain rule of product here for differentiating $ \dfrac{d}{dx}\left( xy \right) $
  & \dfrac{d}{dx}\left( xy \right)=y\times \dfrac{d\left( x \right)}{dx}+x\times \dfrac{d\left( y \right)}{dx} \\
 & \Rightarrow \dfrac{d}{dx}\left( xy \right)=y+\dfrac{xdy}{dx} \\
\[y+\dfrac{xdy}{dx}=\dfrac{d}{dx}\left( {{c}^{2}} \right)\]
Since, $ {{c}^{2}} $ is a constant, so the derivative of this unit x would be zero.
  & \Rightarrow y+x\left( \dfrac{dy}{dx} \right)=0 \\
 & \Rightarrow \dfrac{xdy}{dx}=-y \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-y}{x}\Rightarrow {{m}_{2}}=\dfrac{-y}{x} \\
\[{{m}_{2}}\text{ at point }P({{x}_{1}},{{y}_{1}})\Rightarrow {{m}_{2}}=\dfrac{-{{y}_{1}}}{{{x}_{1}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (2)}\]
From equation (1) and (2) we have:
\[{{m}_{1}}=\dfrac{2a}{{{y}_{1}}}\text{ , }{{\text{m}}_{\text{2}}}=\dfrac{-{{y}_{1}}}{{{x}_{1}}}\]
But \[{{m}_{1}}\times {{\text{m}}_{\text{2}}}=-1\]
\[\Rightarrow \dfrac{2a}{{{y}_{1}}}\text{ }\times \dfrac{-{{y}_{1}}}{{{x}_{1}}}=-1\Rightarrow {{x}_{1}}=2a\]
Putting the value of $ {{x}_{1}} $ from any of the curve to get value of $ {{y}_{1}} $
  & y_{1}^{2}=4a{{x}_{1}}\Rightarrow y=\sqrt{4a{{x}_{1}}}\Rightarrow {{y}_{1}}=\sqrt{4a\times 2a} \\
 & \Rightarrow {{y}_{1}}=\sqrt{8{{a}^{2}}} \\
Now, we have $ {{x}_{1}}=2a\text{ and }{{y}_{1}}=\sqrt{8a} $
Putting values of $ {{x}_{1}}\text{ and }{{\text{y}}_{\text{1}}} $ in curve $ \text{xy=}{{\text{c}}^{\text{2}}} $ we will get the relation between c, a.
So, \[xy={{c}^{2}}\Rightarrow {{x}_{1}}{{y}_{1}}={{c}^{2}}\]
\[\Rightarrow 2a\times \sqrt{8a}={{c}^{2}}\]
Squaring both the sides
  & \Rightarrow {{\left( 2a\times \sqrt{8a} \right)}^{2}}={{c}^{4}} \\
 & \Rightarrow 32{{a}^{4}}={{c}^{4}} \\
Hence proved $ {{c}^{4}}=32{{a}^{4}} $

Note: Geometric representation of the curve is given for better understanding, students must not waste their time imagining the actual scenario instead proceed to the solution as per the orthogonality condition.