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AP EAMCET201922 Apr 2019Morning ShiftMathematicsApplication of DerivativesActual

Tangents are drawn to the curve (y= x ) from the origin. The locus of the points of contact is

Options

  1. A(x y=x+y )
  2. B(x^2 y^2=x^2-y^2 )
  3. C(x y=x-y )
  4. D(x^2 y^2=x^2+y^2 )

Correct answer

B. (x^2 y^2=x^2-y^2 )

Step-by-step solution

Given, (y= x ) Differentiating w.r.t. (x ), we get ( d y d x = x ) If tangent to (y= x ) meet at ((h, k) ) ( ( d y d x )_ (h, k) = h ) ( ) Equation of tangent is ( h(x-h)=y-k ) Since, tangent is passing through (origin). ( h h=k ) But given curve is passing through ((h, k) ) ( aligned k= h & h= 1-k^2 & ^2 h=1-k^2= k^2 h^2 1- k^2 h^2 =k^2 & h^2-k^2=h^2 k^2 aligned ) Hence, locus of point of contact is (x^2-y^2=x^2 y^2 )

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