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

Last updated date: 13th Jul 2024
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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.

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

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

$\text{Hyperbola: yx=}{{\text{c}}^{\text{2}}}$
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{align} & \Rightarrow 2y\dfrac{dy}{dx}=4a \\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}\Rightarrow {{m}_{1}}=\dfrac{2a}{y} \\ \end{align}
${{m}_{1}}\text{ at point }P({{x}_{1}},{{y}_{1}})\Rightarrow {{m}_{1}}=\dfrac{2a}{{{y}_{1}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (1)}$
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{align} & \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{align}
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{align} & \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{align}
${{m}_{2}}\text{ at point }P({{x}_{1}},{{y}_{1}})\Rightarrow {{m}_{2}}=\dfrac{-{{y}_{1}}}{{{x}_{1}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (2)}$
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{align} & 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{align}
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{align} & \Rightarrow {{\left( 2a\times \sqrt{8a} \right)}^{2}}={{c}^{4}} \\ & \Rightarrow 32{{a}^{4}}={{c}^{4}} \\ \end{align}
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.