Let T denote a curve y = y x which is in the first quadrant and let the point 1 ,   0 lie on it. Let the tangent to T at a point P intersect the y-axis at Y P . If P Y P has length 1 for each point P on T , then which of the following is options is/are correct?
D. x y ′ + 1 - x 2 = 0
Let point P is ( h , k ) Now, equation of tangent at P ( y - k ) = m T ( x - h ) ...(i) ( m T = slope of tangent at P ) For Y P , put x = 0 in.....(i) Y P 0 , k - h m T Now, as given P Y P = 1 h 2 + h 2 m T 2 = 1 h 2 1 + m T 2 = 1 1 + m T 2 = 1 h 2 m T 2 = 1 h 2 - 1 m T 2 = 1 - h 2 h 2 m T = ± 1 - h 2 h Put m T = d y d x a n d h = x d y d x = ± 1 - x 2 x ...(i) y = ± - ln 1 + 1 - x 2 x + 1 - x 2 As the curve lies in 1 s t quadrant y must be positive. Hence, y = ln 1 + 1 - x 2 x - 1 - x 2 Also from equation (i) only negative sign will give the correct equation of centre. Hence, d y d x = - 1 - x 2 x x y ′ + 1 - x 2 = 0
Related: Mathematics — Differential Equations · All PYQ Banks