Question

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

Answer 1

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)} \\\ \end{aligned}$$  $$\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$$ Thus, we find the slope of both the tangents through their curve. $ {{T}_{1}} $ corresponds to curve $ {{y}^{2}}=4ax $ Slope $ {{m}_{1}} $ is nothing but $ \dfrac{dy}{dx} $ therefore, differentiating $ {{y}^{2}}=4ax $ with respect of x. $$\begin{aligned} & \Rightarrow 2y\dfrac{dy}{dx}=4a \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}\Rightarrow {{m}_{1}}=\dfrac{2a}{y} \\\ \end{aligned}$$ $${{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)}$$ Now, $ {{T}_{2}} $ corresponds to curve $ \text{xy=}{{\text{c}}^{\text{2}}} $ Therefore, $${{m}_{2}}\Rightarrow \dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( {{c}^{2}} \right)$$ Applying chain rule of product here for differentiating $ \dfrac{d}{dx}\left( xy \right) $ $$\begin{aligned} & \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} \\\ \end{aligned}$$ So, $$y+\dfrac{xdy}{dx}=\dfrac{d}{dx}\left( {{c}^{2}} \right)$$ Since, $ {{c}^{2}} $ is a constant, so the derivative of this unit x would be zero. $$\begin{aligned} & \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} \\\ \end{aligned}$$ $${{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$$ Putting the value of $ {{x}_{1}} $ from any of the curve to get value of $ {{y}_{1}} $ $$\begin{aligned} & 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}}} \\\ \end{aligned}$$ Now, we have $ {{x}_{1}}=2a\text{ and }{{y}_{1}}=\sqrt{8a} $ Putting values of $ {{x}_{1}}\text{ and }{{\text{y}}_{\text{1}}} $ in curve $ \text{xy=}{{\text{c}}^{\text{2}}} $ we will get the relation between c, a. So, $$xy={{c}^{2}}\Rightarrow {{x}_{1}}{{y}_{1}}={{c}^{2}}$$ $$\Rightarrow 2a\times \sqrt{8a}={{c}^{2}}$$ Squaring both the sides $$\begin{aligned} & \Rightarrow {{\left( 2a\times \sqrt{8a} \right)}^{2}}={{c}^{4}} \\\ & \Rightarrow 32{{a}^{4}}={{c}^{4}} \\\ \end{aligned}$$ 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.