AP EAMCET20225 Jul 2022Morning ShiftMathematicsApplication of DerivativesActual
The point on the curve y=x^2+4 x+3 which is closest to the line y=3 x+2 is
Options
- A( 1 2 , 5 4 )
- B( -1 2 , 5 4 )
- C(2, -5 3 )
- D(2, 5 3 )
Correct answer
B. ( -1 2 , 5 4 )
Step-by-step solution
Let (x, y) be on the parabola y=x^2+4 x+3 which is closet to the line y=3 x+2 Perpendicular distance between a point (x₁, y₁ ) and a line (a x+b y+c=0)D= | a x₁+b y₁+c a^2+b^2 | [ . line 3 x-y+2=0 at point . (x₁, y₁ ) ]D= | 3 x-y+2 (3)^2+(-1)^2 |= |3 x- (x^2+4 x+3 )+2 | 9+1 D= |-x^2-x-1 | 10 D= x^2+x+1 10 = (x+ 1 2 )^2+ 3 4 10 On differentiating w.r.t. x , d D d x = (x+ 1 2 )^2 10 + 3 4 10 aligned & d D d x =0, x= -1 2 y=x^2+4 x+3 & y= ( -1 2 )^2+(4) ( -1 2 )+3 y= 1 4 +3-2=0 & y= 1+12-8 4 y= 5 4 aligned d D d x = 2