The prependicular from the origin to the tangent at any pt. on curve is equal to abscissa of that pt Find curve ?

Dear Mayank

let the point on the curve is (x1,y1)

slope of the tangent at this point  is (dy/dx)₁

equation of tangent

y -y1 =(dy/dx)₁ (x-x1)

y -x(dy/dx)₁  +x1(dy/dx)₁  -y1 =0

length of normal from origin

|x1(dy/dx)₁  -y1| /√{(dy/dx)_1²  + 1}  = |x1|

|x1(dy/dx)₁  -y1|   = |x1|√{(dy/dx)_1²  + 1}

square and simplyfy

  {y1² -x1²}/2x1y1  =(dydx)₁

and for general point

 {y² -x²}/2xy  =(dydx)

this is ordinary diferential equation you cab easly solve

 <sup>Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin</sup>