Question

At the point of intersection of rectangular hyperbola $xy={{c}^{2}}$and the parabola ${{y}^{2}}=4ax$tangents to the rectangular hyperbola and the parabola make an angle $\theta$ and $\phi$respectively with the axis of X, then
A. $\theta ={{\tan }^{-1}}\left( -\tan \phi \right)$
B. $\phi ={{\tan }^{-1}}\left( -\tan \theta \right)$
C. $\theta =\dfrac{1}{2}{{\tan }^{-1}}\left( -\tan \phi \right)$
D. $\phi =\dfrac{1}{2}{{\tan }^{-1}}\left( -\tan \theta \right)$

Answer 1

Hint: First you have to find the point of intersection of the rectangular hyperbola $xy={{c}^{2}}$and the parabola ${{y}^{2}}=4ax$ by solving these equations simultaneously.
After that, you need to find the equation of tangent to the hyperbola and the parabola at the point of intersection obtained above.
Complete step by step answer:
Equation of the tangent at point $P\left( {{x}_{1}},{{y}_{1}} \right)$on the rectangular hyperbola $xy={{c}^{2}}$ is $\dfrac{x}{{{x}_{1}}}+\dfrac{y}{{{y}_{1}}}=2$
Equation of tangent at the point $P\left( {{x}_{1}},{{y}_{1}} \right)$ to the parabola ${{y}^{2}}=4ax$is $y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$
After finding the equation of the tangents, convert the both equations in the form $y=mx+c$to get the values of their slopes i.e. “m”.
Now compare the slopes of both the equations of the tangents to get the relation between $\tan \theta$and $\tan \phi$.

Given hyperbola is $xy={{c}^{2}}$ and parabola${{y}^{2}}=4ax$.
Point of intersection of the given rectangular hyperbola and parabola can be obtained.
By simultaneously solving the equations

& xy={{c}^{2}}...............\left( 1 \right)and \\\ & {{y}^{2}}=4ax.............\left( 2 \right) \\\ \end{aligned}$$ From equation (1) we have; $x=\dfrac{{{c}^{2}}}{y}$ Putting this value of x in equation (2), we get $\begin{aligned} & {{y}^{2}}=4a\left( \dfrac{{{c}^{2}}}{y} \right) \\\ & {{y}^{3}}=4a{{c}^{2}} \\\ & y={{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \\\ \end{aligned}$ And now, $xy={{c}^{2}}$ Then $x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}={{c}^{2}}$ $\begin{aligned} & \therefore x=\dfrac{{{c}^{2}}}{{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}} \\\ & x=\dfrac{{{c}^{2}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}{{c}^{\dfrac{2}{3}}}} \\\ & x=\dfrac{{{c}^{2-\dfrac{2}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \\\ & x=\dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \\\ \end{aligned}$ Hence, the required point of intersection is $\left[ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}},{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \right]$(say this point is A) The equation of tangent on the given hyperbola, $xy={{c}^{2}}$at this point of intersection A is; Where, $A=\left[ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}},{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \right]$ $\begin{aligned} & \dfrac{x}{\left\\{ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right\\}}+\dfrac{y}{\left\\{ {{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \right\\}}=2 \\\ & \Rightarrow \dfrac{x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}+y\left( \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right)}{\left\\{ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}}\times {{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \right\\}}=2...............\left( 3 \right) \\\ \end{aligned}$ Since, this point lies on the hyperbola $xy={{c}^{2}}$ Therefore, ${{x}_{1}}{{y}_{1}}={{c}^{2}}$ i.e. $\dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}}\times {{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}={{c}^{2}}$ Now, putting $\dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}}\times {{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}={{c}^{2}}$in equation (3) We get; $\begin{aligned} & \dfrac{x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}+y\left( \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right)}{{{c}^{2}}}=2 \\\ & \Rightarrow x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}+y\left\\{ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right\\}=2{{c}^{2}} \\\ & \Rightarrow x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}+y\left\\{ \dfrac{{{c}^{2}}}{{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}} \right\\}=2{{c}^{2}} \\\ & \Rightarrow x{{\left( 4a{{c}^{2}} \right)}^{\dfrac{2}{3}}}+y{{c}^{2}}=2{{c}^{2}} \\\ & \Rightarrow y{{c}^{2}}=-{{\left( 4a{{c}^{2}} \right)}^{\dfrac{2}{3}}}x+2{{c}^{2}} \\\ \end{aligned}$ Dividing both sides of equation by ${{c}^{2}}$, we get; $y=\dfrac{-{{\left( 4a{{c}^{2}} \right)}^{\dfrac{2}{3}}}}{{{c}^{2}}}x+2$ Comparing this equation with$y=mx+c$, we get; $\begin{aligned} & m=\dfrac{-{{\left( 4a{{c}^{2}} \right)}^{\dfrac{2}{3}}}}{{{c}^{2}}} \\\ & m=\dfrac{-{{\left( 4a \right)}^{\dfrac{2}{3}}}{{c}^{\dfrac{4}{3}}}}{{{c}^{2}}} \\\ & m=-{{\left( 4a \right)}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}} \\\ & m=-{{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{4}{3}}} \\\ & m=-{{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{\left( 2 \right)}^{1}}{{\left( 2 \right)}^{\dfrac{1}{3}}} \\\ & m=-2\left[ {{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}} \right] \\\ \end{aligned}$ Hence, $\tan \theta =-2\left[ {{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}} \right]...................\left( 4 \right)$ Now, equation of tangent at point of intersection A on the parabola $${{y}^{2}}=4ax$$is; Where, $A=\left[ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}},{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}} \right]$ $\begin{aligned} & y{{y}_{1}}=2a\left( x+{{x}_{1}} \right) \\\ & \Rightarrow y{{\left( a{{c}^{2}} \right)}^{\dfrac{1}{3}}}=2a\left[ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}}+x \right] \\\ & \Rightarrow y{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}=2a\left[ \dfrac{{{c}^{\dfrac{4}{3}}}+x{{\left( 4a \right)}^{\dfrac{1}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right] \\\ & \Rightarrow y=2a\left[ \dfrac{{{c}^{\dfrac{4}{3}}}+\left( x \right){{\left( 4a \right)}^{\dfrac{1}{3}}}}{{{\left( 4a \right)}^{\dfrac{2}{3}}}{{\left( c \right)}^{\dfrac{2}{3}}}} \right] \\\ & \Rightarrow y=\dfrac{2a{{\left( 4a \right)}^{\dfrac{1}{3}}}}{{{\left( 4ac \right)}^{\dfrac{2}{3}}}}x+\dfrac{2a{{c}^{\dfrac{4}{3}}}}{{{\left( 4ac \right)}^{\dfrac{2}{3}}}}...............\left( 5 \right) \\\ \end{aligned}$ Comparing equation (5) with $y=m'x+c$ We get $m'=\dfrac{2a{{\left( 4a \right)}^{\dfrac{1}{3}}}}{{{\left( 4ac \right)}^{\dfrac{2}{3}}}}={{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}}$ $\tan \phi =m'={{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}}................\left( 6 \right)$ From equation (4) and (6), we get; $\begin{aligned} & \tan \theta =-2\tan \phi \\\ & \Rightarrow \dfrac{-1}{2}\tan \theta =\tan \phi \\\ & \Rightarrow \dfrac{1}{2}\left( -\tan \theta \right)=\tan \phi \\\ \end{aligned}$ Taking ${{\tan }^{-1}}$on both sides; $\begin{aligned} & \Rightarrow \dfrac{1}{2}{{\tan }^{-1}}\left( -\tan \theta \right)={{\tan }^{-1}}\left( \tan \phi \right) \\\ & \Rightarrow \dfrac{1}{2}{{\tan }^{-1}}\left( -\tan \theta \right)=\phi \\\ \end{aligned}$ Note: This question can also be done easily by finding$\dfrac{dy}{dx}$, which is equal to the slope from the equation of the given curves at the point of intersection. Method II: For hyperbola $xy={{c}^{2}}$ $\Rightarrow y+x\dfrac{dy}{dx}=0$(On differentiating both side with respect to x) $\Rightarrow \dfrac{+dy}{dx}=\dfrac{-y}{x}$ Now, $\dfrac{dy}{dx}$ at point of intersection $A=\tan \theta $ $\begin{aligned} & \tan \theta ={{\left. \dfrac{-y}{x} \right]}_{at\ A}} \\\ & \tan \theta =\dfrac{-{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}}{\left\\{ \dfrac{{{c}^{\dfrac{4}{3}}}}{{{\left( 4a \right)}^{\dfrac{1}{3}}}} \right\\}}=-2\left[ {{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}} \right]...............\left( 7 \right) \\\ \end{aligned}$ Now for parabola, ${{y}^{2}}=4ax$ $\Rightarrow 2y\dfrac{dy}{dx}=4a$ (on differentiating both sides with the respect of x) $\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}=\dfrac{2a}{y}$ At point $A,\dfrac{dy}{dx}$is equal to$\tan \theta $ $\begin{aligned} & \Rightarrow \tan \phi ={{\left. \dfrac{2a}{y} \right]}_{at\ A}} \\\ & \Rightarrow \tan \phi =\dfrac{2a}{{{\left( 4a{{c}^{2}} \right)}^{\dfrac{1}{3}}}}={{a}^{\dfrac{2}{3}}}{{c}^{\dfrac{-2}{3}}}{{2}^{\dfrac{1}{3}}}...............\left( 8 \right) \\\ \end{aligned}$ From equation (7) and (8) we get; $\Rightarrow \dfrac{1}{2}\left( -\tan \theta \right)=\tan \phi $ Taking ${{\tan }^{-1}}$ on both sides; $\theta =\dfrac{1}{2}{{\tan }^{-1}}\left( -\tan \theta \right)$