Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202526 Apr 2025Evening ShiftMathematicsApplication of DerivativesActual

The equation of the tangent to the curve y=b e^ -x / a at the point where it crosses the Y axis is

Options

  1. Ax+ y = ab
  2. Bx a + y ~b =1
  3. Ca x+b y=1
  4. Dx+y=a+b

Correct answer

B. x a + y ~b =1

Step-by-step solution

The curve y = b e^ -x/a crosses the Y-axis at x=0 , giving the point of tangency (0, b) . The derivative is computed as dy dx = - b a e^ -x/a . At x=0 , the slope of the tangent is ( dy dx )_ x=0 = - b a . Using point-slope form with point (0, b) yields y - b = - b a x . Multiplying through by a gives a(y - b) = -b x , rearranged to b x + a y = a b . Dividing both sides by a b results in x a + y b = 1 , which corresponds to option B . The final answer is B .

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