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Home / Doubts and Solutions/JEE/Mathematics

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Asked by rbhatt16 22nd June 2018, 6:48 PM
Answered by Expert
Answer:
begin mathsize 16px style Let space the space curves space intersect space at space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis ax subscript 1 squared plus by subscript 1 squared equals 1 space and space straight a apostrophe straight x subscript 1 squared plus straight b apostrophe straight y subscript 1 squared equals 1 Hence comma space fraction numerator straight x subscript 1 squared over denominator negative straight b plus straight b apostrophe end fraction equals fraction numerator straight y subscript 1 squared over denominator negative straight a plus straight a apostrophe end fraction equals fraction numerator 1 over denominator ab apostrophe minus straight a apostrophe straight b end fraction...... left parenthesis straight I right parenthesis From space first space curve space open parentheses dy over dx close parentheses subscript space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis end subscript equals fraction numerator negative ax subscript 1 over denominator by subscript 1 end fraction and space second space curve space open parentheses dy over dx close parentheses subscript space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis end subscript equals fraction numerator negative straight a apostrophe straight x subscript 1 over denominator straight b apostrophe straight y subscript 1 end fraction According space to space the space question comma fraction numerator negative ax subscript 1 over denominator by subscript 1 end fraction cross times fraction numerator negative straight a apostrophe straight x subscript 1 over denominator straight b apostrophe straight y subscript 1 end fraction equals negative 1 aa apostrophe straight x subscript 1 squared plus bb apostrophe straight y subscript 1 squared equals 0......... left parenthesis II right parenthesis Subtituting space the space values space of space straight x subscript 1 squared space and space straight y subscript 1 squared space from space left parenthesis straight I right parenthesis space and space left parenthesis II right parenthesis solve space it space further space we space get 1 over straight a minus 1 over straight b equals fraction numerator 1 over denominator straight a apostrophe end fraction minus fraction numerator 1 over denominator straight b apostrophe end fraction end style
Answered by Expert 25th June 2018, 10:20 AM
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