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Unformatted text preview: sagbini (ces3287) – HW03 – berg – (55811) 1 This printout should have 18 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 ( 1 , 0) , P 2 (1 , 0) . 1. x 2 4 + y 2 3 = 1 correct 2. x 2 3 + y 2 4 = 1 3. x 2 y 2 3 = 1 4. y 2 x 2 3 = 1 5. x 2 4 y 2 3 = 1 6. x 2 3 + y 2 = 1 7. x 2 + y 2 3 = 1 8. y 2 3 x 2 4 = 1 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 ( 1 , 0) , P 2 (1 , 0) on the xaxis 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 + 1) 2 + y 2 , while dist( Q, P 2 ) = radicalBig ( x 1) 2 + y 2 , so the condition on Q requires that radicalBig ( x + 1) 2 + y 2 + radicalBig ( x 1) 2 + y 2 = 4 . After squaring both sides and simplifying, this becomes ( x + 1) 2 + y 2 + ( x 1) 2 + y 2 + 2 radicalBig (( x + 1) 2 + y 2 )(( x 1) 2 + y 2 ) = 16 . Thus x 2 + y 2 + 1 + radicalBig ( x 2 1) 2 + 2 y 2 ( x 2 + 1) + y 4 = 8 , which after rearranging and squaring gives 7 ( x 2 + y 2 ) = radicalBig ( x 2 + y 2 ) 2 + 2( y 2 x 2 ) + 1 . Hence, after squaring yet again, we obtain 49 14( x 2 + y 2 ) = 2( y 2 x 2 ) + 1 . Consequently, the coordinates x, y of Q sat isfy the equation x 2 4 + y 2 3 = 1 . sagbini (ces3287) – HW03 – berg – (55811) 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 (0 , 2) , P 2 (0 , 2) . 1. x 2 3 + y 2 = 1 2. y 2 3 x 2 4 = 1 3. x 2 3 + y 2 4 = 1 4. x 2 4 + y 2 3 = 1 5. x 2 + y 2 3 = 1 6. y 2 x 2 3 = 1 correct 7. x 2 y 2 3 = 1 8. x 2 4 y 2 3 = 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 (0 , 2) , P 2 (0 , 2) , on the yaxis 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 + ( y + 2) 2 , while dist( Q, P 2 ) = radicalBig x 2 + ( y 2) 2 , so the condition on Q requires that vextendsingle vextendsingle vextendsingle radicalBig x 2 + ( y + 2) 2 radicalBig x 2 + ( y 2) 2 vextendsingle vextendsingle vextendsingle...
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 Spring '11
 Turner
 Conic section

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