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Point on y=x²+7x+2 Nearest to y=3x-3 | JEE Parabola

JEE Maths question with a full step-by-step solution.

Question
The coordinates of the point on the parabola y=x2+7x+2y = x^2 + 7x + 2 which is nearest to the straight line y=3x−3y = 3x - 3 are:
A(−2,−8)(-2, -8)correct
B(1,10)(1, 10)
C(2,20)(2, 20)
D(−1,−4)(-1, -4)
Solution
Step 1: The distance from (x, x2+7x+2)(x,\ x^2 + 7x + 2) to 3x−y−3=03x - y - 3 = 0 is
P=x2+4x+510,P = \frac{x^2 + 4x + 5}{\sqrt{10}},
 the numerator being (x+2)2+1>0(x + 2)^2 + 1 > 0 throughout. Step 2: Minimise: dPdx=2x+410=0⇒x=−2, y=−8\dfrac{dP}{dx} = \dfrac{2x + 4}{\sqrt{10}} = 0 \Rightarrow x = -2,\ y = -8, the point (−2,−8)(-2, -8). Correct answer: (1)
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