Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202311 May 2023Evening ShiftMathematicsApplication of DerivativesActual

The equation of the tangent to the curve y= 9-2 x^2 , at the point where the ordinate and abscissa are equal, is

Options

  1. A2 x+y+ 3 =0
  2. B2 x+y+3 3 =0
  3. C2 x-y-3 3 =0
  4. D2 x+y-3 3 =0

Correct answer

D. 2 x+y-3 3 =0

Step-by-step solution

Given curve is y= 9-2 x^2 If ordinate and abscissa are equal, we get y=x . Equation of the curve becomes x^2=9-2 x^2 x= 3 If x=- 3 , then y= 9-2(3) = 3 In this case, x y . Hence, x - 3 x= 3 and y= 3 Slope of the tangent to the given curve is aligned & 2 y d y ~d x =-4 x & at ( 3 , 3 ), d y ~d x =-2 aligned Equation of the required tangent is aligned & (y- 3 )=-2(x- 3 ) & i.e., 2 x+y-3 3 =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