Examples on Asymptotes and Rectangular Hyperbolas

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Example – 26

Find the equation of the hyperbola with asymptotes   \(3x - 4y + 7 = 0\) and  \(4x + 3y + 1 = 0\) and which passes through the origin.

Solution: The joint equation of the asymptotes is

\[(3x - 4y + 7)(4x + 3y + 1) = 0\]

Since the equation of a hyperbola and that of the joint equation of its asymptotes differs by just a constant, the equation of the hyperbola must be

\[(3x - 4y + 7)(4x + 3y + 1) + k = 0\]

We can obtain k by exploiting the fact that this hyperbola passes through the origin (0, 0) ; thus

\[\begin{align} & \qquad\quad(7)(1) + k = 0 \\ &  \Rightarrow \qquad  k = - 7 \end{align} \]

The equation of the hyperbola is

\[\begin{align}  &\qquad\quad (3x - 4y + 7)(4x + 3y + 1) - 7 = 0 \\\\ & \Rightarrow \qquad  12{x^2} - 7xy - 12{y^2} + 31x + 17y = 0 \end{align} \]

Example - 27

A circle cuts the rectangular hyperbola   \(xy = 1\)  in four points  \(({x_i},\,{y_i}),\,i = 1,\;2,\,3,\,4.\) Prove that

\[{x_1}\,{x_2}\,{x_3}\,{x_4} = {y_1}\,{y_2}\,{y_3}\,{y_4} = 1\]

Solution: We assume the equation of the circle to be

\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]

Using \(\begin{align}y = \frac{1}{x}\end{align}\)  in this equation, we have

\[\begin{align}&{x^2} + \frac{1}{{{x^2}}} + 2gx + \frac{{2f}}{x} + c = 0 \\  &{x^4} + 2g{x^3} + c{x^2} + 2fx + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)  \\ \end{align} \]

This is a biquadratic in x yielding four roots \({x_1},\,{x_2},\,{x_3},\,{x_4}.\)

It is evident from (1) that

\[{x_1}\,{x_2}\,{x_3}\,{x_4} = 1\]

Since \(\begin{align}{y_i} = \frac{1}{{{x_i}}},\end{align}\)  the result   \({y_1}\,{y_2}\,{y_3}\,{y_4} = 1\) follows.

Example - 28

A variable line of slope 4 intersects  \(xy = 1\) in A and B. C divides AB internally in the ratio 1 : 2. Find the locus of C.

Solution: Let A and B have the coordinates  \(\left( {{t_1},\,\begin{align}\frac{1}{{{t_1}}}\end{align}} \right)\) and  \(\left( {{t_2},\,\begin{align}\frac{1}{{{t_2}}}\end{align}}\right).\)

Since AB has slope 4, we have

\[\begin{align}& \frac{{\begin{align}\frac{1}{{{t_1}}} - \frac{1}{{{t_2}}}\end{align}}}{{{t_1} - {t_2}}} = 4 \\  \Rightarrow\quad & {t_1}{t_2} = \frac{{ - 1}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \end{align} \]

Assume the coordinates of C to be  \((h,\,k).\)  We thus have

\[\begin{align}&\qquad\quad h = \frac{{2{t_1} + {t_2}}}{3},\,\,\,\,\,k = \frac{\begin{align}{\frac{2}{{{t_1}}} + \frac{1}{{{t_2}}}}\end{align}}{3} \\   &\Rightarrow\quad   2{t_1} + {t_2} = 3h,\,\,\,\,\,2{t_2} + {t_1} = 3k{t_1}{t_2}  \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;=  - \frac{{3k}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Using{\text{ }}\left( 1 \right)} \right)  \\
   &\Rightarrow\quad   3({t_1} + {t_2}) = 3h - \frac{{3k}}{4},\,\,\,\,\,({t_1} - {t_2}) = 3h + \frac{{3k}}{4}  \\
   &\Rightarrow\quad   {t_1} + {t_2} = h - \frac{k}{4},\,\,\,\,\,{t_1} - {t_2} = 3\left( {h + \frac{k}{4}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)  \\ 
\end{align} \]

Using  \({\left( {\,{t_1} + {t_2}} \right)^2} = {\left( {{t_1} - {t_2}} \right)^2} + 4{t_1}{t_2}\)  and the value of  \({t_1}\,{t_2}\)   from (1), we can eliminate \({t_1}\) and  \({t_2}\) from (2) to obtain a relation in h and k :

\[16{h^2} + 10hk + {k^2} - 2 = 0\]

Thus, the locus of C is

\[16{x^2} + 10xy + {y^2} - 2 = 0\]

Example - 29

On any point P on the hyperbola   \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,\end{align}\) a tangent is drawn intersecting the asymptotes in B and C. Let O be the centre of the hyperbola. Prove that

(a) area  \((\Delta OBC)\) is constant           (b)  \(PB = PC\) 

Solution:

Assume P to be the point   \((a\sec \theta ,\,b\tan \theta ).\)  The equation of the tangent BPC is

\[\frac{x}{a}\sec \theta  - \frac{y}{b}\tan \theta  = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\]

The two asymptotes have the equation

\[y =  \pm \frac{b}{a}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

The point B and C can therefore be determined by simultaneously solving (1) and (2):

\[B \equiv (a(\sec \theta  + \tan \theta ),\,\,b(\sec \theta  + \tan \theta )),\,\,\,\,\,C \equiv (a(\sec \theta  - \tan \theta ),\,\, - b(\sec \theta  - \tan \theta ))\]

The area of   \(\Delta OBC\)  can be evaluated using the determinant formula :

\[\Delta  = \frac{1}{2}\left| {\begin{array}{*{20}{c}}  {\,\,a(\sec \theta  + \tan \theta )}&{b(\sec \theta  + \tan \theta )}&{1\,\,} \\   {\,a(\sec \theta  - \tan \theta )}&{ - b(\sec \theta  - \tan \theta )}&{1\,\,} \\ 
  0&0&{1\,\,} \end{array}} \right|\]

Expanding along \({R_3},\) we have

\[\begin{align}&\Delta  = \frac{1}{2}\left( {2ab\left( {{{\sec }^2}\theta  - {{\tan }^2}\theta } \right)} \right) \ \\  \,\,\,\,\, &\;\;\;= ab  \\\end{align} \]

which is constant.

That P is the mid-point of BC is straight away evident by simple inspection of the coordinates of B and C.

Example - 30

Four points A, B, C, D lie on a rectangular hyperbola  \(xy = {c^2}\) such that   \(AC \bot BD.\)  Let O be the the centre of the hyperbola. The slopes of  \(OA,\,OB,\,OC\) and OD are  \({m_1},\,{m_2},\,{m_3}\) and  \({m_4}\) respectively. Prove that \({m_1}\,{m_2}\,{m_3}\,{m_4} = 1.\)

Solution:

The coordinates of A, B,C, D can be assumed to be  \(\left( {c{t_i},\,\begin{align}\frac{c}{{{t_i}}}\end{align}} \right),\,i = 1,\,2,\,3,\,4.\) Thus,

\[\begin{align}&\qquad\;\;\; {{m}_{i}}=\frac{\begin{align}\frac{c}{{{t}_{i}}}\end{align}-0}{c{{t}_{i}}-0}=\frac{1}{t_{i}^{2}} \\ &\Rightarrow \quad{{m}_{1}}\,{{m}_{2}}\,{{m}_{3}}\,{{m}_{4}}=\frac{1}{t_{1}^{2}t_{2}^{2}t_{3}^{2}t_{4}^{2}}\qquad\qquad\qquad\qquad\ldots \left( 1 \right) \\ 
\end{align}\]

Since  \(AC\bot BD,\) we have

\[\begin{align} &\qquad \frac{\begin{align}\frac{c}{{{t}_{3}}}-\frac{c}{{{t}_{1}}}\end{align}}{c{{t}_{3}}-c{{t}_{1}}}\times \frac{\begin{align}\frac{c}{{{t}_{2}}}-\frac{c}{{{t}_{4}}}\end{align}}{c{{t}_{2}}-c{{t}_{4}}}=-1 \\ & \Rightarrow\quad {{t}_{1}}\,{{t}_{2}}\,{{t}_{3}}\,{{t}_{4}}=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 2 \right) \end{align}\]

Using (1) and (2), we have

\[{{m}_{1}}\,{{m}_{2}}\,{{m}_{3}}\,{{m}_{4}}=1\]