Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202523 Apr 2025Morning ShiftMathematicsApplication of DerivativesActual

The point on the curve 4 y^2-4 y+2 x-1=0 at which the tangent becomes parallel toY-axis is

Options

  1. A(1, 1 2 )
  2. B( 1 2 , 1 )
  3. C(-1,- 1 2 )
  4. D( 1 2 , 0 )

Correct answer

A. (1, 1 2 )

Step-by-step solution

The condition for a tangent to be parallel to the Y-axis is that dx dy = 0 . Differentiate 4y^2 - 4y + 2x - 1 = 0 with respect to y : 8y - 4 + 2 dx dy = 0 Solving for dx dy gives dx dy = 2 - 4y . Set dx dy = 0 , so 2 - 4y = 0 , yielding y = 1 2 . Substitute into the original equation: 4 ( 1 2 )^2 - 4 ( 1 2 ) + 2x - 1 = 0 1 - 2 + 2x - 1 = 0 2x - 2 = 0 , so x = 1 . The point is (1, 1 2 ) , corresponding to option A .

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025If f(x)=x e^ x(1-x) , x R , then f(x) is 2025The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All MHT CET PYQs