Question

If , for some prove that is a constant independent of a and b .

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Solution

It is given that ( x−a ) 2 + ( y−b ) 2 = c 2 . We have to prove that [ 1+ ( dy dx ) 2 ] 3 2 d 2 y d x 2 is a constant independent of aand b. Let the equation ( x−a ) 2 + ( y−b ) 2 = c 2 and differentiate it with respect to x. d dx { ( x−a ) 2 + ( y−b ) 2 }= d dx ( c 2 ) 2 ( x−a ) ( 2−1 ) d dx ( x−a )+2 ( y−b ) ( 2−1 ) d dx ( y−b )=0 2( x−a )+2( y−b )( dy dx )=0 dy dx = −( x−a ) ( y−b ) Differentiate the above equation with respect to x. d dx ( dy dx )= d dx { −( x−a ) ( y−b ) } d 2 y d x 2 =[ ( y−b ) d dx ( x−a )− d dx ( y−b )×( x−a ) ( y−b ) 2 ] =[ ( y−b )×1−( dy dx )×( x−a ) ( y−b ) 2 ] =[ ( y−b )×1− ( −x+a ) ( y−b ) ×( x−a ) ( y−b ) 2 ] Further simplify. d 2 y d x 2 =[ ( y−b ) 2 + ( x−a ) 2 ( y−b ) 2 ×( y−b ) ] d 2 y d x 2 = − c 2 ( y−b ) 3 Substitute the value of dy dx , d 2 y d x 2 in [ 1+ ( dy dx ) 2 ] 3 2 d 2 y d x 2 . [ 1+ ( dy dx ) 2 ] 3 2 d 2 y d x 2 = [ 1+ { −( x−a ) ( y−b ) } 2 ] 3 2 { − c 2 ( y−b ) 3 } =− [ ( y−b ) 2 + ( x−a ) 2 ( y−b ) 2 { c 2 ( y−b ) 3 } ] 3 2 =− [ c 2 ( y−b ) 2 ] 3 2 { c 2 ( y−b ) 3 } Further simplify. [ 1+ ( dy dx ) 2 ] 3 2 d 2 y d x 2 =− [ c 2 ( y−b ) 2 ] 3 2 × ( y−b ) 3 c 2 =− ( c y−b ) 2× 3 2 × ( y−b ) 3 c 2 =−( c 3 c 2 )× ( y−b ) 3 ( y−b ) 3 =−c Hence the required condition proved.

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