JEE Main2007MathematicsApplication of DerivativesActual

The normal to a curve at P(x, y) meets the x -axis at G . If the distance of G from the origin is twice the abscissa of P , then the curve is a

Options

Aellipse
Bparabola
Ccircle
Dpair of straight lines

Correct answer

A. ellipse

Step-by-step solution

Equation of normal is Y-y=- d x d y (X-x) aligned & G (x+y d y d x , 0 ) & |x+y d y d x |=|2 x| & y d y d x =x or y d y d x =-3 x & y d y=x d x or y d y=-3 x d x & y^2 2 = x^2 2 +c or y^2 2 =- 3 x^2 2 +c & x^2-y^2=-2 c or 3 x^2+y^2=2 c. aligned

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the function f: (0, ) (- , ) given by f(x) = x , _e(x) - x + 1 . Then which one of the following statements is TRUE? 2026 Let P be the point on the parabola y = x^2 such that the slope of the tangent to the parabola at the point P is 4 . Let Q be the point in the first quadrant lying on the circle x^2 2026 Let f: R R be a differentiable function such that f ( x+y 3 ) = f(x) + f(y) 3 for all x, y R , and f'(0) = 3 . Then the minimum value of the function g(x) = 3 + e^x f(x) , is: 2026 _ 0 x (16 ( x 2 ) ^3 ( x 2 ) ) is equal to: 2026 Let f(x) be a polynomial of degree 5 , and have extrema at x = 1 and x = -1 . If _ x 0 ( f(x) x^3 ) = -5 , then f(2) - f(-2) is equal to: 2026 The number of critical points of the function f(x) = cases | x x |, & x 0 1, & x = 0 cases in the interval (-2 , 2 ) is equal to : 2026 Let f be a differentiable function satisfying f(x)=1-2 x+ ₀^ x e ^ (x-t) f(t) dt , x R and let g (x)= ₀^ x (f( t )+2)¹⁵( t -4)⁶( t +12)¹⁷ dt , x R . If p and q are respectively the 2026 Consider the following three statements for the function f:(0, ) R defined by f(x)= | _ e x |-|x-1| : (I) f is differentiable at all x>0 . (II) f is increasing in (0,1) . (III) f i 2026 Full Application of Derivatives list All JEE Main PYQs