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AP EAMCET202318 May 2023Evening ShiftMathematicsApplication of DerivativesActual

The locus of the point on the curve y= x where the tangent drawn at that point always passes through the point (0, ) is

Options

  1. Ax = y -
  2. Bx+ y+1=0
  3. Cx ^2 (1- y ^2 )=( y - )^2
  4. Dx ^2+( y - )^2=0

Correct answer

C. x ^2 (1- y ^2 )=( y - )^2

Step-by-step solution

y= x y^ = x Equation of tangent line at (x₁, y₁ ) : (y-y₁ )= x₁ (x-x₁ ) Eq ^ n ( i ) passes through the point (0, ) aligned & ( -y₁ )= x₁ (0-x₁ ) & ( -y₁ )= 1-y₁^2 (-x₁ ) & ( -y₁ )^2= (1-y₁^2 ) x₁^2 & x₁ (1-y₁^2 )= (y₁- )^2 aligned Taking locus of point (x₁, y₁ ) , we get x(1-y)^2=(y- )^2

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