# Hyperbolas Set 2

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Example - 3

For any triangle inscribed in a rectangular hyperbola, prove that its orthocentre lies on the hyperbola.

Solution: Assume the hyperbola to be  $$xy = {c^2}$$  and assume any three points  $$A({t_1}),\,B({t_2})$$  and  $$C({t_3})$$ on it.

The equation of AB is

\begin{align}&\quad\;\frac{{y - \begin{align}\frac{c}{{{t_2}}}\end{align}}}{{x - c{t_2}}} = \frac{{\begin{align}\frac{c}{{{t_2}}} - \frac{c}{{{t_1}}}\end{align}}}{{c{t_2} - c{t_1}}} \\\\ &\Rightarrow x + y{t_1}{t_2} = c({t_1} + {t_2}) \end{align}

Similarly, the equations of BC and CA are

\begin{align} BC:x + y\,{t_2}{t_3} = c({t_2} + {t_3}) \\\\ CA:x + y\,{t_3}{t_1} = c({t_3} + {t_1}) \\ \end{align}

We now find any two of the altitudes, say AD and BE, and find their point of intersection O :

.\begin{align} AD:y - x{t_2}{t_3} = \frac{c}{{{t_1}}} - c{t_1}{t_2}{t_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ BE:y - x{t_1}{t_3} = \frac{c}{{{t_2}}} - c{t_1}{t_2}{t_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right) \\\end{align}

The point of intersection O of (1) and (2) can easily be found :

$O \equiv \left( { - \frac{c}{{{t_1}{t_2}{t_3}}},\, - c{t_1}{t_2}{t_3}} \right)$

From the coordinates of O, it should be clear that O lies on the hyperbola.

Example - 4

Show that an ellipse and a hyperbola having the same foci intersect orthogonally in four distinct points.

Solution: We can assume the equations of the ellipse and the hyperbola as follows:

\begin{align} &\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Ellipse} \right) \\\\ &\frac{{{x^2}}}{{{c^2}}} - \frac{{{y^2}}}{{{d^2}}} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Hyperbola} \right) \\ \end{align}

If we assume $${{e}_{1}}\,\text{and}\,{{e}_{2}}$$ to be their respective eccentricities, we have :

\begin{align}&\quad\qquad{b^2} = {a^2}(1 - e_1^2) \\\\ &\quad\qquad {d^2} = {c^2}(e_2^2 - 1) \\\\ &\quad\qquad a{e_1} = c{e_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(due to the foci being these same})\ \\\\ & \Rightarrow \quad{d^2} = {c^2}\left( {\frac{{{a^2}e_1^2}}{{{c^2}}} - 1} \right) \\\\ & \,\,\,\,\,\,\, \qquad\;= {a^2}e_1^2 - {c^2} \\\\ &\Rightarrow\quad {c^2} + {d^2} = {a^2}e_1^2 = {a^2} - {b^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left....( 1 \right)\\ \end{align}

To evaluate the angle of intersection of the two curves, we must evaluate their respective derivatives at the point of intersection.

By symmetry, it should be evident that the angle of intersection at all the four points will the same; thus we can focus our attention on any one point of intersection, say A.

By solving the equations for the hyperbola simultaneously, A can be obtained:

$A \equiv \left( {\frac{{ac\sqrt {{b^2} + {d^2}} }}{{\sqrt {{b^2}{c^2} + {a^2}{d^2}} }},\,\,\frac{{bd\sqrt {{a^2} - {c^2}} }}{{\sqrt {{b^2}{c^2} + {a^2}{d^2}} }}} \right) = \left( {\alpha ,\,\beta } \right)$

\left\{ \begin{align} &\alpha \,{\text{and }}\beta {\text{ are being}} \\ & {\text{introduced just}} \\ & {\text{for convenence}} \\ \end{align} \right\}_{}^{}

At A, the ellipse has a derivative given by

${m_e} = {\left. {\frac{{dy}}{{dx}}} \right|_{A(\alpha ,\,\beta )}} = {\left. {\frac{{ - {b^2}x}}{{{a^2}y}}} \right|_{(\alpha ,\,\beta )}} = \frac{{ - {b^2}\alpha }}{{{a^2}\beta }}$

Similarly, at A, the hyperbola has a derivative given by

${m_h} = {\left. {\frac{{dy}}{{dx}}} \right|_{\Delta (\alpha ,\,\beta )}} = {\left. {\frac{{{d^2}x}}{{{c^2}y}}} \right|_{(\alpha ,\,\beta )}} = \frac{{{d^2}\alpha }}{{{c^2}\beta }}$

Now, we evaluate $${m_e}\,{m_h}:$$

\begin{align}&{m_e}\,{m_h} = \frac{{ - {b^2}{d^2}}}{{{a^2}{c^2}}} \times \frac{{{\alpha ^2}}}{{{\beta ^2}}}\\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;= \frac{{ - {b^2}{d^2}}}{{{a^2}{c^2}}} \times \frac{{{a^2}{c^2}}}{{{b^2}{d^2}}} \times \frac{{{b^2} + {d^2}}}{{{a^2} - {c^2}}} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;= 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Using{\text{ }}\left( 1 \right)} \right) \\ \end{align}

which implies that the two curves intersect orthogonally.

Example - 5

A variable circle passing through the origin touches the hyperbola  $$xy = {c^2}$$ at P and cuts the hyperbola at two other distinct points Q and R. Determine the locus of the foot of the perpendicular X drawn from the origin O upon QR.

Solution: We can assume point P to be  \left( {ct,\,\begin{align}\frac{c}{t}\end{align}} \right).  At point P, the hyperbola and the circle have a common tangent given by the equation

\begin{align}&\qquad\;\;\;{L_1}:T\left( {ct,\,\frac{c}{t}} \right) = 0 \\ & \Rightarrow\quad {L_1}:x + y{t^2} = 2ct \\ \end{align}

We can assume QR to be the line

${L_2}:x\cos \theta + y\sin \theta = p$

Why we have assumed in normal form will soon become clear.

We will assume a family of circles approach. Any second degree curve passing through P, Q and R can be written as

\begin{align}&\qquad\;\;\;\ {L_1}{L_2} + \lambda (xy - {c^2}) = 0 \\\\&\Rightarrow\quad (x + y{t^2} - 2ct)(x\cos \theta + y\sin \theta - p) + \lambda (xy - {c^2}) = 0 \\ \end{align}

This represents a circle passing through the origin only if:

\begin{align}&\text{Co eff. of}\; {x^2} = \text{Co eff. of }{y^2}\qquad\qquad \quad \cos \theta = \sin \theta \cdot {t^2} \qquad\qquad\qquad\quad\ldots \left( 1 \right) \\\\ &\text{Co eff. of}\;\; xy = 0 \qquad\qquad\qquad\qquad\qquad \lambda + \sin \theta + {t^2}\cos \theta = 0\qquad\qquad\ldots \left( 2 \right) \\\\ &\text{Constant term} = 0 \qquad\qquad\qquad\qquad\quad 2tp = c\lambda \qquad\qquad\qquad\quad\quad\quad\quad\ldots \left( 3 \right) \end{align}

We need to find the locus of   $$X(h,\,k),$$   the foot of the perpendicular dropped upon  $${L_2}$$  from the origin O.

The advantage of assuming  $${L_2}$$ in normal form will now be evident : h and k, the coordinates of X, satisfy the simple relations

\begin{align}&h = p\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 4 \right) \\& k = p\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 5 \right)\\ \end{align}

From (1) and (2), we have

\begin{align}&\qquad\;\;\lambda + \sin \theta + \frac{{{{\cos }^2}\theta }}{{\sin \theta }} = 0 \\ \\ &\Rightarrow\quad \lambda \sin \theta + 1 = 0\qquad\qquad\qquad \ldots \left( 6 \right) \end{align}

From (3),

\begin{align}&\qquad\;\;\;{\lambda ^2} = \frac{{4{p^2}{t^2}}}{{{c^2}}} \\ \\ &\Rightarrow \quad\frac{{4{p^2}}}{{{c^2}}} \times \frac{{\cos \theta }}{{\sin \theta }} = \frac{1}{{{{\sin }^2}\theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {From{\text{ }}\left( 6 \right){\text{ }}and{\text{ }}\left( 1 \right)} \right) \\\\ &\Rightarrow \quad \frac{{4{p^2}}}{{{c^2}}}\sin \theta \cos \theta = 1 \\\\ &\Rightarrow \quad \frac{{4{p^2}}}{{{c^2}}} \cdot \frac{k}{p} \cdot \frac{h}{p} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {From{\text{ }}\left( 4 \right){\text{ }}and{\text{ }}\left( 5 \right)} \right) \\ \\ &\Rightarrow \quad hk = \frac{{{c^2}}}{4} \end{align}

Thus, the locus of X is another rectangular hyperbola xy = \begin{align}\frac{{{c^2}}}{4}.\end{align}

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