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If (x/a sin A - y/b cosA)=1 and ( x/a cosA + y/b sinA)=1, then find the value of (x^2/a^2+y^2/b^2). |
| Answer» Given: {tex}\\frac { x } { a } \\cos \\theta + \\frac { y } { b } \\sin \\theta = 1 \\text { and } \\frac { x } { a } \\sin \\theta - \\frac { y } { b } \\cos \\theta = 1 , {/tex}To prove: {tex}\\frac { x ^ { 2 } } { a ^ { 2 } } + \\frac { y ^ { 2 } } { b ^ { 2 } } = 2{/tex}Now, {tex}{\\left( {\\frac{x}{a}\\cos \\theta + \\frac{y}{b}\\sin \\theta } \\right)^2} + {\\left( {\\frac{x}{a}\\sin \\theta - \\frac{y}{b}\\cos \\theta } \\right)^2} = {(1)^2} + {(1)^2}{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}{\\cos ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\sin ^2}\\theta + 2\\frac{x}{a}\\cos \\theta \\frac{y}{b}\\sin \\theta + \\frac{{{x^2}}}{{{a^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\cos ^2}\\theta - 2\\frac{x}{a}\\sin \\theta \\frac{y}{b}\\cos \\theta = 1 + 1{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}{\\cos ^2}\\theta + \\frac{{{x^2}}}{{{a^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\cos ^2}\\theta = 2{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}\\left( {{{\\cos }^2}\\theta + {{\\sin }^2}\\theta } \\right) + \\frac{{{y^2}}}{{{b^2}}}\\left( {{{\\sin }^2}\\theta + {{\\cos }^2}\\theta } \\right) = 2{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}} + \\frac{{{y^2}}}{{{b^2}}} = 2{/tex}\xa0{tex}\\left[ {\\because {{\\cos }^2}\\theta + {{\\sin }^2}\\theta = 1} \\right]{/tex}Hence proved. | |