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CBSE Class 11-science Answered

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Asked by araima2001 | 14 Mar, 2017, 08:08: AM
answered-by-expert Expert Answer
begin mathsize 16px style The space equation space of space the space tangent space at space open parentheses 4 cosθ comma space fraction numerator 16 over denominator square root of 11 end fraction sinθ close parentheses space to space the space ellipse space 16 straight x squared plus 11 straight y squared equals 256 space is space given space by 16 open parentheses 4 cosθ close parentheses straight x plus 11 open parentheses fraction numerator 16 over denominator square root of 11 end fraction sinθ close parentheses straight y equals 256 Divide space throughout space by space 16 comma space we space get 4 cosθx plus fraction numerator 11 over denominator square root of 11 end fraction sinθy equals 16 rightwards double arrow 4 cosθx plus fraction numerator 11 square root of 11 over denominator square root of 11 square root of 11 end fraction sinθy equals 16 rightwards double arrow 4 cosθx plus square root of 11 sinθy equals 16 Since space the space tangent space is space also space straight a space tangent space to space the space circle space straight x squared plus straight y squared minus 2 straight x equals 15 comma that space is comma space straight x squared minus 2 straight x plus 1 plus straight y squared equals 15 plus 1 rightwards double arrow space left parenthesis straight x minus 1 right parenthesis squared plus straight y squared equals 16 rightwards double arrow space left parenthesis straight x minus 1 right parenthesis squared plus straight y squared equals 4 squared We space know space that space distance space of space straight a space tangent space from space the space centre space of space the space circle space will space be space equal space to space its space radius comma space that space is comma space 4 space units. Since space that space distance space will space be space the space perpendicular comma So comma space fraction numerator vertical line 4 cosθ minus 16 vertical line over denominator square root of 16 cos squared straight theta plus 11 sin squared straight theta end root end fraction equals 4 rightwards double arrow fraction numerator 4 vertical line cosθ minus 4 vertical line over denominator square root of 16 cos squared straight theta plus 11 left parenthesis 1 minus cos squared straight theta right parenthesis end root end fraction equals 4 rightwards double arrow fraction numerator vertical line cosθ minus 4 vertical line over denominator square root of 16 cos squared straight theta plus 11 left parenthesis 1 minus cos squared straight theta right parenthesis end root end fraction equals 1 Squaring space both space sides comma space we space get fraction numerator open parentheses cosθ minus 4 close parentheses squared over denominator 16 cos squared straight theta plus 11 left parenthesis 1 minus cos squared straight theta right parenthesis end fraction equals 1 rightwards double arrow fraction numerator open parentheses cosθ minus 4 close parentheses squared over denominator 16 cos squared straight theta plus 11 minus 11 cos squared straight theta end fraction equals 1 rightwards double arrow fraction numerator open parentheses cosθ minus 4 close parentheses squared over denominator 5 cos squared straight theta plus 11 end fraction equals 1 rightwards double arrow cos squared straight theta minus 8 cosθ plus 16 equals 5 cos squared straight theta plus 11 rightwards double arrow 4 cos squared straight theta minus 5 plus 8 cosθ equals 0 rightwards double arrow 4 cos squared straight theta plus 8 cosθ minus 5 equals 0 rightwards double arrow 4 cos squared straight theta plus 10 cosθ minus 2 cosθ minus 5 equals 0 rightwards double arrow 2 cosθ left parenthesis 2 cosθ plus 5 right parenthesis minus left parenthesis 2 cosθ plus 5 right parenthesis equals 0 rightwards double arrow left parenthesis 2 cosθ plus 5 right parenthesis left parenthesis 2 cosθ minus 1 right parenthesis equals 0 rightwards double arrow 2 cosθ plus 5 equals 0 space space space or space space space 2 cosθ minus 1 equals 0 rightwards double arrow cosθ equals fraction numerator negative 5 over denominator 2 end fraction space space space space or space space space space space cosθ equals 1 half space space Now comma space we space know space that space minus 1 less or equal than cosθ less or equal than 1 comma space so space cosθ equals fraction numerator negative 5 over denominator 2 end fraction space is space not space possible. Hence comma space cosθ equals 1 half space space rightwards double arrow straight theta equals cos to the power of negative 1 end exponent open parentheses 1 half close parentheses rightwards double arrow straight theta equals plus-or-minus straight pi over 3 end style
Dear student, there is no 'K' in the question. I am guessing you mean find the value of θ. Please get back to us if you have any query regarding this.
Answered by Rebecca Fernandes | 14 Mar, 2017, 09:03: AM

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