Question

The tangent to the hyperbola $xy={{c}^{2}}$ at a point P intersects the x-axis at T and the y-axis at T’. The normal to the curve at the same point intersects the x-axis at N and the y-axis at T’. If $\Delta$ is the area of the triangle PNT and $\Delta '$ is the area of the triangle PN’T’, then $\dfrac{1}{\Delta }+\dfrac{1}{\Delta '}$ is
[a] Equal to 1
[b] Depends on the point taken
[c] Depends on c
[d] Equal to 2.

Answer 1

Hint: Assume that the point P on the curve $xy={{c}^{2}}$ be $P\left( x,\dfrac{{{c}^{2}}}{x} \right)$. Find the slope of the tangent to the curve at Point P by differentiating y with respect to x. Hence find the equation of the tangent and hence find the coordinates of points T and T’. Use the fact that the tangent and normal are perpendicular to each other. Hence find the equation of the normal and hence find the coordinates of the points N and N’. Hence find the areas of the triangle PNT and PN’T’ and hence find the value of $\dfrac{1}{\Delta }+\dfrac{1}{\Delta '}$

Complete step by step solution:
Let the $P\left( {{x}_{1}},{{y}_{1}} \right)$be a point on the curve $xy={{c}^{2}}$
Satifying the point on the curve, we have
$\begin{aligned} & {{x}_{1}}{{y}_{1}}={{c}^{2}} \\\ & \Rightarrow {{y}_{1}}=\dfrac{{{c}^{2}}}{{{x}_{1}}} \\\ \end{aligned}$
Hence, we have
$P\equiv \left( {{x}_{1}},\dfrac{{{c}^{2}}}{{{x}_{1}}} \right)$
Now, we have
$xy={{c}^{2}}\Rightarrow y=\dfrac{{{c}^{2}}}{x}$
Differentiating both sides with respect to x to get the slope of the tangent at point (x,y) on the curve, we get
$\dfrac{dy}{dx}=\dfrac{-{{c}^{2}}}{{{x}^{2}}}$
Hence, we have the slope of the tangent at point P $={{\left. \dfrac{dy}{dx} \right|}_{x={{x}_{1}},y={{y}_{1}}}}=\dfrac{-{{c}^{2}}}{{{x}_{1}}^{2}}$
Hence, we have
Equation of tangent using point slope form of equation of a line is
$y-\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{-{{c}^{2}}}{{{x}_{1}}^{2}}\left( x-{{x }_{1}} \right)\text{ }\left( a \right)$
The points T and T’ lie on this line
For point T’, we have x = 0 since the point T’ lies on the y-axis.
Hence if we substitute x = 0 in the equation (a), we will get the y coordinate of point T’.
Substituting x = 0 in the equation (a), we get
$y-\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{-{{c}^{2}}}{{{x}_{1}}^{2}}\left( -{{x}_{1}} \right)=\dfrac{{{c}^{2}}}{{{x}_{1}}}$
Adding $\dfrac{{{c}^{2}}}{{{x}_{1}}}$on both sides, we get
$y=\dfrac{2{{c}^{2}}}{{{x}_{1}}}$
Hence, we have
$T'\equiv \left( 0,\dfrac{2{{c}^{2}}}{{{x}_{1}}} \right)$
For point T, we have y = 0 since the point lies on the x-axis.
Hence if we substitute y = 0 in the equation (a), we will get the x-coordinate of the point T.
Substituting y = 0 in the equation (a), we get
$-\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{-{{c}^{2}}}{{{x}_{1}}^{2}}\left( x-{{x}_{1}} \right)$
Multiplying both sides by $\dfrac{-x_{1}^{2}}{{{c}^{2}}}$, we get
${{x}_{1}}=x-{{x}_{1}}$
Adding ${{x}_{1}}$ on both sides, we get
$x=2{{x}_{1}}$
Hence, we have
$T\equiv \left( 2{{x}_{1}},0 \right)$
Let m be the slope of the normal.
We know that if the slope of two perpendicular lines are ${{m}_{1}}$ and ${{m}_{2}}$, then ${{m}_{1}}{{m}_{2}}=-1$
Hence, we have
$m\times \left( \dfrac{-{{c}^{2}}}{{{x}_{1}}^{2}} \right)=-1$
Multiplying both sides by $\dfrac{-x_{1}^{2}}{{{c}^{2}}}$, we get
$m=\dfrac{x_{1}^{2}}{{{c}^{2}}}$
Hence, we have
$y-\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{x_{1}^{2}}{{{c}^{2}}}\left( x-{{x}_{1}} \right)\text{ }\left( b \right)$
The points N and N’ lie on this line.
For point N’, we have x = 0, since the point lies on the y-axis.
Hence if we substitute x = 0 in equation b, we will get the y-coordinate of the point N’.
Substituting x = 0 in equation (b), we get
$\begin{aligned} & y-\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{x_{1}^{2}}{{{c}^{2}}}\left( -{{x}_{1}} \right) \\\ & \Rightarrow y=\dfrac{{{c}^{2}}}{{{x}_{1}}}-\dfrac{x_{1}^{3}}{{{c}^{2}}}=\dfrac{{{c}^{4}}-x_{1}^{4}}{{{c}^{2}}{{x}_{1}}} \\\ \end{aligned}$
Hence, we have
$N'\equiv \left( 0,\dfrac{{{c}^{4}}-x_{1}^{4}}{{{c}^{2}}{{x}_{1}}} \right)$
For point N, we have y = 0, since the point lies on the x-axis.
Hence if we substitute y = 0 in equation (b), we will get the x-coordinate of the point N.
Substituting y = 0 in equation (b), we get
$\begin{aligned} & -\dfrac{{{c}^{2}}}{{{x}_{1}}}=\dfrac{x_{1}^{2}}{{{c}^{2}}}\left( x-{{x}_{1}} \right) \\\ & \Rightarrow x={{x}_{1}}-\dfrac{{{c}^{4}}}{x_{1}^{3}}=\dfrac{x_{1}^{4}-{{c}^{4}}}{x_{1}^{3}} \\\ \end{aligned}$
Hence, we have
$N\equiv \left( \dfrac{x_{1}^{4}-{{c}^{4}}}{x_{1}^{3}},0 \right)$
Now, we know that the area of the triangle formed by the points $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ is given by
$A=\dfrac{1}{2}\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} \\\ \end{matrix} \right|$
For triangle PNT, we have
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2{{x}_{1}},0 \right),\left( {{x}_{3}},{{y}_{3}} \right)=\left( \dfrac{x_{1}^{4}-{{c}^{4}}}{x_{1}^{3}},0 \right)$
Hence, we have
$\begin{aligned} & \Delta =\dfrac{1}{2}\left| \begin{matrix} 2{{x}_{1}}-{{x}_{1}} & 0-{{y}_{1}} \\\ \dfrac{x_{1}^{4}-{{c}^{4}}}{x_{1}^{3}}-{{x}_{1}} & 0-{{y}_{1}} \\\ \end{matrix} \right|=\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & -{{y}_{1}} \\\ \dfrac{-{{c}^{4}}}{x_{1}^{3}} & -{{y}_{1}} \\\ \end{matrix} \right| \\\ & =\dfrac{1}{2}\left| {{x}_{1}}{{y}_{1}}+\dfrac{{{c}^{4}}}{x_{1}^{3}}{{y}_{1}} \right|=\dfrac{1}{2}\left| {{c}^{2}}+\dfrac{{{c}^{6}}}{x_{1}^{4}} \right|=\dfrac{{{c}^{2}}}{2x_{1}^{4}}\left( x_{1}^{4}+{{c}^{4}} \right) \\\ \end{aligned}$
Similarly for triangle PN’T’, we have
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},\dfrac{{{c}^{2}}}{{{x}_{1}}} \right),\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,\dfrac{2{{c}^{2}}}{{{x}_{1}}} \right),\left( {{x}_{3}},{{y}_{3}} \right)=\left( 0,\dfrac{{{c}^{4}}-x_{1}^{4}}{{{c}^{2}}{{x}_{1}}} \right)$
Hence, we haves
$\begin{aligned} & \Delta '=\dfrac{1}{2}\left| \begin{matrix} 0-{{x}_{1}} & \dfrac{2{{c}^{2}}}{{{x}_{1}}}-\dfrac{{{c}^{2}}}{{{x}_{1}}} \\\ 0-{{x}_{1}} & \dfrac{{{c}^{4}}-x_{1}^{4}}{{{c}^{2}}{{x}_{1}}}-\dfrac{{{c}^{2}}}{{{x}_{1}}} \\\ \end{matrix} \right|=\dfrac{1}{2}\left| \begin{matrix} -{{x}_{1}} & \dfrac{{{c}^{2}}}{{{x}_{1}}} \\\ -{{x}_{1}} & \dfrac{-x_{1}^{4}}{{{c}^{2}}{{x}_{1}}} \\\ \end{matrix} \right| \\\ & =\dfrac{1}{2}\left| \dfrac{x_{1}^{4}}{{{c}^{2}}}+{{c}^{2}} \right|=\dfrac{1}{2{{c}^{2}}}\left( {{c}^{4}}+x_{1}^{4} \right) \\\ \end{aligned}$
Hence, we have
$\dfrac{1}{\Delta }+\dfrac{1}{\Delta '}=\dfrac{2x_{1}^{4}}{{{c}^{2}}\left( x_{1}^{4}+{{c}^{4}} \right)}+\dfrac{2{{c}^{2}}}{x_{1}^{4}+{{c}^{4}}}=\dfrac{2}{{{c}^{2}}\left( x_{1}^{4}+{{c}^{4}} \right)}\left( x_{1}^{4}+{{c}^{4}} \right)=\dfrac{2}{{{c}^{2}}}$
Hence option [c] is correct.
Note: Alternatively, we can solve the question using the parametric form of the equation of hyperbola $xy={{c}^{2}}$ as $x=ct$ and $y=\dfrac{c}{t}$, where t is a parameter. This method will make the above calculations easier.