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CBSE Class 11-science Answered

<div>If the tangent at the point P(x1, y1) to the parabola y²=4ax meets the parabola y² = 4a(x+b) at Q and R, then the midpoint of QR is  ?</div>

Asked by Malavika Umesh | 20 May, 2015, 05:49: PM

Expert Answer

g i v e n space t h a t space P open parentheses x subscript 1 comma y subscript 1 close parentheses space b e space a n y space p o i n t space t o space t h e space p a r a b o l a space y squared equals 4 a x T h u s comma space t h e space tan g e n t space t o space t h e space p a r a b o l a space a t space t h i s space p o i n t space i s space y y subscript 1 equals 2 a open parentheses x plus x subscript 1 close parentheses... left parenthesis 1 right parenthesis S i n c e space open parentheses x subscript 1 comma y subscript 1 close parentheses space l i e s space o n space t h e space p a r a b o l a comma space w e space h a v e comma space space y subscript 1 squared equals 4 a x subscript 1... open parentheses 2 close parentheses L e t space Q open parentheses x subscript 2 comma y subscript 2 close parentheses space a n d space R open parentheses x subscript 3 comma y subscript 3 close parentheses space b e space a n y space t w o space p o i n t s space o n space t h e space p a r a b o l a comma space y squared equals 4 a open parentheses x plus b close parentheses N o w space c o n s i d e r space t h e space e q u a t i o n space o f space t h e space p a r a b o l a space y squared equals 4 a open parentheses x plus b close parentheses M u l t i p l y space y subscript 1 squared cross times y squared equals 4 a space y subscript 1 squared open parentheses x plus b close parentheses R e w r i t i n g space t h e space a b o v e space e q u a t i o n comma space w e space h a v e comma 4 a squared open parentheses x plus x subscript 1 close parentheses squared equals 4 a space y subscript 1 squared open parentheses x plus b close parentheses space space space space space space space space left square bracket f r o m space e q u a t i o n space left parenthesis 1 right parenthesis right square bracket rightwards double arrow a open parentheses x plus x subscript 1 close parentheses squared equals 4 a x subscript 1 open parentheses x plus b close parentheses space space space space space space space space space space space space space space left square bracket f r o m space e q u a t i o n space left parenthesis 2 right parenthesis right square bracket rightwards double arrow a open parentheses x squared plus x subscript 1 squared plus 2 x x subscript 1 close parentheses equals 4 a x x subscript 1 plus 4 a b x subscript 1 rightwards double arrow a x squared minus 2 a x x subscript 1 plus a x subscript 1 squared minus 4 a b x subscript 1 equals 0 rightwards double arrow x squared minus 2 x x subscript 1 plus x subscript 1 squared minus 4 b x subscript 1 equals 0 T h e space r o o t s space o f space t h e space a b o v e space e q u a t i o n space a r e space t h e space p o i n t s space Q space a n d space R T h u s comma space s u m space o f space t h e space r o o t s comma x subscript 2 plus x subscript 3 equals 2 x subscript 1 rightwards double arrow fraction numerator x subscript 2 plus x subscript 3 over denominator 2 end fraction equals x subscript 1 T h u s comma space t h e space x minus c o o r d i n a t e space o f space t h e space m i d p o i n t space o f space Q space a n d space R space i s space x subscript 1 S o comma space w e space n e e d space t o space f i n d space t h e space y minus c o o r d i n a t e space o f space t h e space m i d p o i n t space o f space Q space a n d space R. T h e space e q u a t i o n space o f space t h e space tan g e n t space i s space y y subscript 1 equals 2 a open parentheses x plus x subscript 1 close parentheses S u b s t i t u t i n g space x equals x subscript 1 comma space w e space h a v e comma space y y subscript 1 equals 2 a open parentheses x subscript 1 plus x subscript 1 close parentheses rightwards double arrow space y y subscript 1 equals 2 a open parentheses x subscript 1 plus x subscript 1 close parentheses rightwards double arrow space y y subscript 1 equals 4 a x subscript 1 rightwards double arrow space y equals fraction numerator 4 a x subscript 1 over denominator y subscript 1 end fraction T h u s space t h e space m i d p o i n t space o f space Q space a n d space R space i s space open square brackets x subscript 1 comma fraction numerator 4 a x subscript 1 over denominator y subscript 1 end fraction close square brackets

Answered by Vimala Ramamurthy | 21 May, 2015, 10:22: AM

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