Question

If the curves ${{y}^{2}}=4ax$ and $xy={{c}^{2}}$ cut orthogonally then $\dfrac{{{c}^{4}}}{{{a}^{4}}}=$ $A.4$
B.8$C.16$
D.32$$$$

Answer 1

We denote the point of intersection of the curves as $\left( {{x}_{1}},{{y}_{1}} \right)$. We differentiate equation curve ${{y}^{2}}=4ax$ with respect to $x$and find the slope of tangent at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ as ${{m}_{1}}$. We differentiate equation curve $xy={{c}^{2}}$ with respect to $x$and find the slope of tangent at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ as ${{m}_{2}}$. We use conditions on slopes for perpendicular lines and have ${{m}_{1}}{{m}_{2}}=-1$. We find ${{x}_{1}}$ in terms $a$ which we put in ${{y}^{2}}=4ax$ to get ${{y}_{1}}$. We put ${{x}_{1}},{{y}_{1}}$ in $xy={{c}^{2}}$ and find $\dfrac{{{c}^{4}}}{{{a}^{4}}}$.$$$$

Complete step-by-step answer:
We know from differential calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point.
We are given in the question equations of the following curves.

& {{y}^{2}}=4ax.......\left( 1 \right) \\\ & xy={{c}^{2}}..........\left( 2 \right) \\\ \end{aligned}$$ Let us assume the point of intersection of curve (1) and curve (2) be$\left( {{x}_{1}},{{y}_{1}} \right)$. Let us differentiate the equation of curve (1) with respect to $x$ and find the slope at any point. We have; $$\begin{aligned} & \dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4ax \right) \\\ & \Rightarrow 2y\dfrac{dy}{dx}=4a \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}=\dfrac{2a}{y} \\\ \end{aligned}$$ So the slope of tangent of curve (1) at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is say ${{m}_{1}}$then we have; $${{m}_{1}}={{\left. \dfrac{dy}{dx} \right|}_{x={{x}_{1}},y={{y}_{1}}}}=\dfrac{2a}{{{y}_{1}}}$$ $\left( {{x}_{1}},{{y}_{1}} \right)$. Let us differentiate the equation of curve (2) with respect to $x$and find the slope at any point. We have; $$\begin{aligned} & \dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( {{c}^{2}} \right) \\\ & \Rightarrow x\dfrac{dy}{dx}+y=0 \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-y}{x} \\\ \end{aligned}$$ So the slope of tangent of curve (1) at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is say ${{m}_{2}}$then we have; $${{m}_{2}}={{\left. \dfrac{dy}{dx} \right|}_{x={{x}_{1}},y={{y}_{1}}}}=\dfrac{-{{y}_{1}}}{{{x}_{1}}}$$ We are given the curve cut each other orthogonally which means their tangents will be perpendicular to each other at the point of intersection. We know that the product of slope of two perpendicular lines is $-1.$ So at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ we have; $$\begin{aligned} & {{m}_{1}}\times {{m}_{2}}=-1 \\\ & \Rightarrow \dfrac{2a}{{{y}_{1}}}\times \dfrac{-{{y}_{1}}}{{{x}_{1}}}=-1 \\\ & \Rightarrow {{x}_{1}}=2a \\\ \end{aligned}$$ Let us put ${{x}_{1}}=2a$ in the equation of curve (1) since $\left( {{x}_{1}},{{y}_{1}} \right)$ lies on curve (1) and have; $$\begin{aligned} & {{y}^{2}}=4\times a\times \left( 2a \right) \\\ & \Rightarrow {{y}^{2}}=8{{a}^{2}} \\\ & \Rightarrow y=\pm 2\sqrt{2}a \\\ \end{aligned}$$ We put the value of ${{x}_{1}},{{y}_{1}}$in equation of curve (2) since $\left( {{x}_{1}},{{y}_{1}} \right)$ also lies on curve (2) and have; $$\begin{aligned} & xy={{c}^{2}} \\\ & \Rightarrow 2a\times \left( \pm 2\sqrt{2}a \right)={{c}^{2}} \\\ & \Rightarrow \pm 4\sqrt{2}{{a}^{2}}={{c}^{2}} \\\ & \Rightarrow \dfrac{{{c}^{2}}}{{{a}^{2}}}=\pm 4\sqrt{2} \\\ \end{aligned}$$ We square both sides to have the required result as $$\Rightarrow \dfrac{{{c}^{4}}}{{{a}^{4}}}={{\left( \pm 4\sqrt{2} \right)}^{4}}=32$$ So the correct option is D.$$$$  **Note:** We note that given equation ${{y}^{2}}=4ax$ is an equation of parabola opened right wards which has tangent equation at $\left( {{x}_{1}},{{y}_{1}} \right)$ is $y{{y}_{1}}=4a\left( x+{{x}_{1}} \right)$ and equation $xy={{c}^{2}}$is an equation of rectangular hyperbola which tangent equation at $\left( {{x}_{1}},{{y}_{1}} \right)$ is $x{{y}_{1}}+y{{x}_{1}}=2{{c}^{2}}$. The equations are given in implicit form. We should differentiate them implicitly rather than expressing tem in the explicit form $y=f\left( x \right)$ and then differentiating and making it difficult. We note that orthogonal is the general term for perpendicular which includes vectors, curves and all mathematical objects while perpendicular is used for lines.