Courses for Kids
Free study material
Free LIVE classes
Join Vedantu’s FREE Mastercalss

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

335.1k+ views
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.
seo images

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

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

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.
Last updated date: 23rd Sep 2023
Total views: 335.1k
Views today: 7.35k