Examples on Normals To Hyperbolas Set 1

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Example - 17

Find the condition that must be satisfied if the line  \(px + qy + r = 0\) is to be a normal to the hyperbola\(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}\)

Solution : Any normal to the given hyperbola has the form

\[ax\sin \theta  + by = ({a^2} + {b^2})\tan \theta \]

If

\[px + qy =  - r\]

is to be a normal to this hyperbola, we must have some value of \(\theta \) such that

\[\begin{align}&\qquad\;\;\frac{{a\sin \theta }}{p} = \frac{b}{q} = \frac{{({a^2} + {b^2})\tan \theta }}{{ - r}}  \\
   &\Rightarrow  \quad {\text{cosec}}\,\theta  = \frac{{aq}}{{bp}} \;\;\;\;\;\;;\cot \theta  =  - \frac{{({a^2} + {b^2})q}}{{br}} \\ \end{align} \]

\(\theta \) can now easily be eliminated using the relation  \({\text{cose}}{{\text{c}}^2}\theta  - {\cot ^2}\theta  = 1\) to obtain

\[\begin{align}&\qquad\;\;\;\;\;\frac{{{a^2}{q^2}}}{{{b^2}{p^2}}} - \frac{{{{({a^2} + {b^2})}^2}{q^2}}}{{{b^2}{r^2}}} = 1  \\&\Rightarrow \qquad  \frac{{{a^2}}}{{{p^2}}} - \frac{{{b^2}}}{{{q^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{r^2}}}  \\\end{align}\]

This is the requisite condition.

Example - 18

If the normal at the four points \(({x_i},\,{y_i}),\,\,i = 1,\,2,\,3,\,4\) on the hyperbola  \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) are concurrent, show that

\[\left( {\sum {{x_i}} } \right)\left( {\sum {\frac{1}{{{x_i}}}} } \right) = \left( {\sum {{y_i}} } \right)\left( {\sum {\frac{1}{{{y_i}}}} } \right) = 4\]

Solution : The nature of the problem might be able to give you a hint that we must try to form a biquadratic equation in  \(x(\text{or}\,\,y)\) to obtain the desired result.

Let the point of concurrency of the four normals be\(P(h,\,\,k).\)

Now, the normal to the hyperbola at \((X,\,\,Y)\) has the form

\[\frac{{{a^2}x}}{X} + \frac{{{b^2}y}}{Y} = {a^2} + {b^2}\]

If this passes through P, we have

\[\frac{{{a^2}h}}{X} + \frac{{{b^2}k}}{Y} = {a^2} + {b^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]

(1) must be satisfied by four values of  \((X,\,\,Y),\) namely  \(({x_i},\,\,{y_i}),\,\,i = 1,\,2,\,3,\,4.\) Since  \((X,\,\,Y)\) lie on the hyperbola, we have

\[\frac{{{X^2}}}{{{a^2}}} - \frac{{{Y^2}}}{{{b^2}}} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

From (1) and (2), it should be evident that we can form a biquadratic equation in either X or Y, by eliminating the other. Let us form a biquadratic in X.

From (1), we have

\[\begin{align}&\frac{{{b^2}k}}{Y} = {a^2} + {b^2} - \frac{{{a^2}h}}{X}  \\\\   \,\,\,\,\,\,\,\, &\qquad=\frac{{({a^2} + {b^2})X - {a^2}h}}{X}  \\\\ &\Rightarrow \quad  Y = \frac{{{b^2}kX}}{{({a^2} + {b^2})X - {a^2}h}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)  \\ \end{align} \]

Using (3) in (2), we have

\[\begin{align} &\qquad\qquad \frac{{{X^2}}}{{{a^2}}} - \frac{{{b^4}{k^2}{X^2}}}{{{b^2}{{\{ ({a^2} + {b^2})X - {a^2}h\} }^2}}} = 1 \\  & \Rightarrow  \quad  {(\lambda X - {a^2}h)^2}{X^2} - {a^2}{b^2}{k^2}{X^2} = {a^2}{\{ \lambda X - {a^2}h\} ^2}\left[ \begin{gathered}  {\text{where }}\lambda  = {a^2} + {b^2}{\text{ has }}  \\  {\text{been substituted for }}  \\  {\text{convenience}}  \\\end{gathered}  \right]  \\ &  \Rightarrow   \quad {\lambda ^2}{X^4} - 2{a^2}h\lambda {X^3} + ({a^4}{h^2} - {a^2}{b^2}{k^2} - {a^2}{\lambda ^2}){X^2} + 2{a^4}h\lambda X - {a^6}{h^2} = 0 \\\end{align} \]

This biquadratic yields four X values, namely\({x_1},\,{x_2},\,{x_3},\,{x_4}.\)

We have

\[\begin{align}\sum {{x_i} = \frac{{2{a^2}h\lambda }}{{{\lambda ^2}}} = \frac{{2{a^2}h}}{\lambda }}   \\\\\sum {\frac{1}{{{x_i}}} = \frac{{2{a^4}h\lambda }}{{{a^6}{h^2}}} = \frac{{2\lambda }}{{{a^2}h}}} \\\end{align} \]

This immediately yields

\[\left( {\sum {{x_i}} } \right)\left( {\sum {\frac{1}{{{x_i}}}} } \right) = 4\]

We can analogously prove the other equality.