MHT CET202617 April 2026Morning ShiftMathematicsApplication of DerivativesActual
The tangent to the curve intersects the Y-axis at point P. A line drawn through point P is perpendicular to this tangent and passes through another point (1, 0) . The differential equation of the curve is...
Options
- Ay dy dx - x ( dy dx )^2 = 1
- Bx dy dx - y ( dy dx )^2 = 1
- Cy dy dx + x = 1
- Dx dy dx + y = 1
Correct answer
A. y dy dx - x ( dy dx )^2 = 1
Step-by-step solution
Let the point on the curve be (x, y) . The equation of the tangent at (x, y) is Y - y = dy dx (X - x) . To find the intersection with the Y-axis (point P), substitute X = 0 : Y = y - x dy dx Thus, the coordinates of point P are (0, y - x dy dx ) . The slope of the tangent is dy dx , so the slope of the line perpendicular to it is - dx dy . This perpendicular line passes through P (0, y - x dy dx ) and (1, 0) . The slope of this line is also given by: 0 - (y - x dy dx ) 1 - 0 = x dy dx - y Equating the two expressio