PrepRung
Sign inSign up free

JEE Main 2025 · Mathematics · Differentiation and Application of Derivatives · medium

The least value of the square of the distance of a point of the parabola y=x^2 from the point (0,3) is

Single correct option

The least value of the square of the distance of a point of the parabola y=x2y=x^2 from the point (0,3)(0,3) is
  1. A. 33
  2. B. 52\dfrac52
  3. C. 94\dfrac94
  4. D. 114\dfrac{11}{4}
Show answer and solution

Answer: (D) 114\dfrac{11}{4}

For a point (x,x2)(x,x^2) the squared distance from (0,3)(0,3) is D=x2+(x2−3)2D=x^2+(x^2-3)^2. Put u=x2≥0u=x^2\ge0, so D(u)=u+(u−3)2=u2−5u+9D(u)=u+(u-3)^2=u^2-5u+9. Then D′(u)=2u−5=0D'(u)=2u-5=0 gives u=52u=\dfrac52, which is admissible since u≥0u\ge0, and D′′=2>0D''=2>0 shows a minimum. Therefore Dmin⁡=254−252+9=114D_{\min}=\dfrac{25}4-\dfrac{25}2+9=\dfrac{11}4. The two nearest points are (±5/2, 5/2)\left(\pm\sqrt{5/2},\,5/2\right). Checking the boundary u=0u=0 gives D=9D=9, which is larger. Because the minimising abscissa satisfies x2=52x^2=\tfrac52, the nearest points on the parabola are (±5/2,52)\left(\pm\sqrt{5/2},\tfrac52\right).
Report a mistake

Original question written in the JEE Main 2025 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake