Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Consider the hyperbola with equation $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 .$ Let F be the focus with coordinates $(\sqrt{A^{2} + B^{2}}, 0) .$ Let $e = \frac{\sqrt{A^{2} + B^{2}}}{A}$ and l be the vertical line with equation $x = \frac{A}{e} .$ Show that for any point P on the hyperbola, $\frac{F P}{D P} = e,$ where D is the point on l closest to P.

The given statement is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

The given hyperbola equation is $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 .$ and the coordinates of focus are $(\sqrt{A^{2} + B^{2}}, 0) .$

Step 2. Showing.

Let P be the point $(x, y)$ on the hyperbola.

So, the distance between the points $(x, y) (\sqrt{A^{2} + B^{2}}, 0) .$

By using the formula for the distance $\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} .$

$Here, x_{1} = x, y_{1} = y, x_{2} = \sqrt{A^{2} - B^{2}}, y_{2} = 0 F P = \sqrt{{(x - (\sqrt{A^{2} - B^{2}}))}^{2} + (y - 0)^{2}} F P = \sqrt{(x^{2} + A^{2} - B^{2} - 2 x \sqrt{A^{2} - B^{2}}) + y^{2}} F P = \sqrt{(x^{2} + A^{2} - B^{2} - 2 x \sqrt{A^{2} - B^{2}}) - B^{2} + \frac{B^{2} x^{2}}{A^{2}}} [\sin c e, \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 \Rightarrow y^{2} = \frac{B^{2} x^{2}}{A^{2}} - B^{2}] F P = \sqrt{x^{2} (1 + \frac{B^{2}}{A^{2}}) + A^{2} - 2 x \sqrt{A^{2} + B^{2}}} F P = \sqrt{(\frac{A^{2} + B^{2}}{A^{2}}) (x^{2} - 2 x \sqrt{A^{2} + B^{2}} \frac{A^{2}}{A^{2} + B^{2}} + A^{2} \cdot \frac{A^{2}}{A^{2} + B^{2}})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} \sqrt{(x^{2} - 2 x \sqrt{A^{2} + B^{2}} \frac{A^{2}}{{(\sqrt{A^{2} + B^{2}})}^{2}} + A^{2} \cdot \frac{A^{2}}{{(\sqrt{A^{2} + B^{2}})}^{2}})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} \sqrt{(x^{2} - 2 x \frac{A^{2}}{(\sqrt{A^{2} + B^{2}})} + {(\frac{A^{2}}{(\sqrt{A^{2} + B^{2}})})}^{2})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} (x - \frac{A^{2}}{\sqrt{A^{2} + B^{2}}}) F P = e (x - \frac{A}{e}) [since \frac{\sqrt{A^{2} + B^{2}}}{A} = e] \frac{F P}{D P} = \frac{e (x - \frac{A}{e})}{x - \frac{A}{e}} [\sin c e F P = e (x - \frac{A}{e}), D P = x - \frac{A}{e}] \frac{F P}{D P} = e$ uncaught exception: Invalid chunkin file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('d5a92cf1d81b51b...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage(' $" width="0" height="0" role="math" style="max-width: none;" localid="1649555159482"> Here, x_{1} = x, y_{1} = y, x_{2} = \sqrt{A^{2} + B^{2}}, y_{2} = 0 F P = \sqrt{{(x - (\sqrt{A^{2} + B^{2}}))}^{2} + (y - 0)^{2}} F P = \sqrt{(x^{2} + A^{2} + B^{2} - 2 x \sqrt{A^{2} + B^{2}}) + y^{2}} F P = \sqrt{(x^{2} + A^{2} + B^{2} - 2 x \sqrt{A^{2} + B^{2}}) - B^{2} + \frac{B^{2} x^{2}}{A^{2}}} [\sin c e \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 \Rightarrow y^{2} = \frac{B^{2} x^{2}}{A^{2}} - B^{2}] F P = \sqrt{x^{2} (1 + \frac{B^{2}}{A^{2}}) + A^{2} - 2 x \sqrt{A^{2} + B^{2}}}$

Step 3. Showing.

By proceeding with the calculation further,

$F P = \sqrt{(\frac{A^{2} + B^{2}}{A^{2}}) (x^{2} - 2 x \sqrt{A^{2} - B^{2}} \frac{A^{2}}{A^{2} + B^{2}} + A^{2} \cdot \frac{A^{2}}{A^{2} + B^{2}})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} \sqrt{(x^{2} - 2 x \sqrt{A^{2} + B^{2}} \frac{A^{2}}{{(\sqrt{A^{2} + B^{2}})}^{2}} + A^{2} \cdot \frac{A^{2}}{{(\sqrt{A^{2} + B^{2}})}^{2}})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} \sqrt{(x^{2} - 2 x \frac{A^{2}}{(\sqrt{A^{2} + B^{2}})} + {(\frac{A^{2}}{(\sqrt{A^{2} + B^{2}})})}^{2})} F P = \sqrt{\frac{A^{2} + B^{2}}{A^{2}}} (x - \frac{A^{2}}{\sqrt{A^{2} + B^{2}}}) F P = e (x - \frac{A}{e}) [since \frac{\sqrt{A^{2} + B^{2}}}{A} = e] Now, \frac{F P}{D P} = \frac{e (x - \frac{A}{e})}{x - \frac{A}{e}} [\sin c e F P = e (x - \frac{A}{e}), D P = x - \frac{A}{e}] \frac{F P}{D P} = e$

Hence proved.

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