# 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

${{m}_{1}}\,{{m}_{2}}\,{{m}_{3}}\,{{m}_{4}}=1$

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