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AP EAMCET202317 May 2023Evening ShiftMathematicsApplication of DerivativesActual

If a normal drawn at a point P to the curve y= x passes through the origin, then the locus of P is

Options

  1. Ax ^2= y ^2- y ^4
  2. Bx+y=1
  3. C1 y^2 - 1 x^2 =1
  4. D1 y^4 - 1 x^4 =1

Correct answer

A. x ^2= y ^2- y ^4

Step-by-step solution

y= x d y d x = x Slope of normal m=- 1 d y d x =- 1 x Let the co-ordinate of point P is (h, k) Then, m=- 1 h Equation of normal, which passes through (0,0) is (y-0)=m(x-0) y=- 1 h x Above equation passes through (h, k) k=- h h h=- h k ...(i) Also, (h, k) lies on y= x k= h h=k ..(ii) Equation (i) ^2+( ii )^2 : aligned & ^2 h+ ^2 h= h^2 k^2 +k^2 & 1= h^2+k^4 k^2 h^2+k^4=k^2 aligned Taking locus of point (h, k) , we get aligned & x^2+y^4=y^2 & x^2=y^2-y^4 aligned

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