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Unformatted text preview: qazi (kaq87) – HW03 – berg – (55290) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the relation between x and y when the point Q ( x, y ) has the property that dist( Q, P 1 ) + dist( Q, P 2 ) = 4 with respect to points P 1 (0 , − 1) , P 2 (0 , 1) . 1. y 2 3 − x 2 4 = 1 2. x 2 − y 2 3 = 1 3. x 2 4 + y 2 3 = 1 4. x 2 4 − y 2 3 = 1 5. x 2 + y 2 3 = 1 6. x 2 3 + y 2 = 1 7. y 2 − x 2 3 = 1 8. x 2 3 + y 2 4 = 1 correct Explanation: The distance condition dist( Q, P 1 ) + dist( Q, P 2 ) = 4 ensures that Q traces out an ellipse with foci at P 1 (0 , − 1) , P 2 (0 , 1) on the yaxis and so has graph P 1 P 2 Q The question really asks for the equation of this ellipse. First notice that dist( Q, P 1 ) = radicalBig x 2 + ( y + 1) 2 , while dist( Q, P 2 ) = radicalBig x 2 + ( y − 1) 2 , so the condition on Q requires that radicalBig x 2 + ( y + 1) 2 + radicalBig x 2 + ( y − 1) 2 = 4 . After squaring both sides and simplifying, this becomes x 2 + ( y + 1) 2 + x 2 + ( y − 1) 2 + 2 radicalBig ( x 2 + ( y + 1) 2 )( x 2 + ( y − 1) 2 ) = 16 . Thus x 2 + y 2 + 1 + radicalBig x 4 + 2 x 2 ( y 2 + 1) + ( y 2 − 1) 2 = 8 , which after rearranging and squaring gives 7 − ( x 2 + y 2 ) = radicalBig ( x 2 + y 2 ) 2 + 2( x 2 − y 2 ) + 1 . Hence, after squaring yet again, we obtain 49 − 14( x 2 + y 2 ) = 2( x 2 − y 2 ) + 1 . Consequently, the coordinates x, y of Q sat isfy the equation x 2 3 + y 2 4 = 1 . qazi (kaq87) – HW03 – berg – (55290) 2 keywords: conic section, ellipse, hyperbola, distance definition, distance, 002 10.0 points Find the relation between x and y when the point Q ( x, y ) has the property that  dist( Q, P 1 ) − dist( Q, P 2 )  = 2 with respect to points P 1 ( − 2 , 0) , P 2 (2 , 0) . 1. y 2 − x 2 3 = 1 2. x 2 4 + y 2 3 = 1 3. x 2 3 + y 2 4 = 1 4. x 2 4 − y 2 3 = 1 5. y 2 3 − x 2 4 = 1 6. x 2 + y 2 3 = 1 7. x 2 − y 2 3 = 1 correct 8. x 2 3 + y 2 = 1 Explanation: The distance condition  dist( Q, P 1 ) − dist( Q, P 2 )  = 2 ensures that Q traces out a hyperbola with foci at P 1 ( − 2 , 0) , P 2 (2 , 0) on the xaxis and so has graph P 1 P 2 Q The question really asks for the equation of this hyperbola. First notice that dist( Q, P 1 ) = radicalBig ( x + 2) 2 + y 2 , while dist( Q, P 2 ) = radicalBig ( x − 2) 2 + y 2 , so the condition on Q requires that vextendsingle vextendsingle vextendsingle radicalBig ( x + 2) 2 + y 2 − radicalBig ( x − 2) 2 + y 2 vextendsingle vextendsingle vextendsingle = 2 . After squaring both sides and simplifying, this becomes ( x + 2) 2 + y 2 + ( x − 2) 2 + y 2 − 2 radicalBig (( x + 2) 2 + y 2 )(( x − 2) 2 + y 2 ) = 4 ....
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This note was uploaded on 03/28/2012 for the course M 408 M taught by Professor Gilbert during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Gilbert
 Multivariable Calculus

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