MHT CET202410 May 2024Morning ShiftMathematicsApplication of DerivativesActual

The curve x^4-2 x y^2+y^2+3 x-3 y=0 cuts the X -axis at (0,0) at an angle of

Options

Correct answer

A. 4

Step-by-step solution

Given equation of curve is x^4-2 x y^2+y^2+3 x-3 y=0...(i) Slope of tangent =m= d y d x at (0,0) Differentiating (i) w.r.to x , we get array ll & 4 x^3-2 x 2 y ~d y ~d x -y^2(2)+2 y ~d y ~d x +3-3 ~d y ~d x =0 & ~d y ~d x = 4 x^3-2 y^2+3 4 x y-2 y+3 & ~m = . d y ~d x |_ (0,0) = 4(0)-2(0)+3 4(0) (0)-2(0)+3 = 3 3 =1 & ~m =1 & =1 & = 4 array

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 2025 The 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 2025 If f(x)=x e^ x(1-x) , x R , then f(x) is 2025 The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025 If 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 = 2025 If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025 If 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 = 2025 If 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