Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202312 May 2023Evening ShiftMathematicsApplication of DerivativesActual

The equation of the normal to the curve 3 x^2-y^2=8 , which is parallel to the line x+3 y=10 , is

Options

  1. Ax+3 y+6=0
  2. Bx+3 y-3=0
  3. Cx+3 y+8=0
  4. Dx+3 y-4=0

Correct answer

C. x+3 y+8=0

Step-by-step solution

3 x^2-y^2=8 Differentiating w.r.t. x , we get aligned & 6 x-2 y d y ~d x =0 & d y ~d x = 3 x y aligned Slope of the tangent to the curve is 3 x y . Slope of the normal is -y 3 x . It is parallel to line x+3 y=10 slope =- 1 3 -y 3 x = -1 3 x=y When x=y , equation of the curve becomes 3 x^2-x^2=8 x^2=4 x=2,-2 y=2,-2 (2,2) and (-2,-2) are the points of contact of the normal and the curve. Equations are (y-2)= -1 3 (x-2) or aligned & (y+2)= -1 3 (x+2) & i.e., x+3 y-8=0 or x+3 y+8=0 aligned

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