Application of IntegralsmediumFree

Application of Integrals: Let Point Origin Lying Parabola Normal Line Parabola

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

Question
Let PP be a point (not the origin) lying on the parabola x2=yx^2 = y. The normal line to the parabola at PP will intersect the parabola at another point QQ. The minimum possible value for the area bounded by the line PQPQ and the parabola is
A23\dfrac23
B34\dfrac34
C43\dfrac43correct
D35\dfrac35
Solution
Question attachment Step 1: Let P=(t,t2)P = \left(t,t^2\right). PP is not the origin, so t≠0t \ne 0, and the parabola is symmetric about the yy-axis, so let t>0t>0. Since dydx=2x\dfrac{dy}{dx} = 2x, the tangent slope at PP is 2t2t and the normal slope is −12t-\dfrac{1}{2t}:
y−t2=−12t(x−t)⟹x+2ty=t+2t3y-t^2 = -\frac{1}{2t}\left(x-t\right) \quad\Longrightarrow\quad x+2ty = t+2t^3
 Step 2: Putting y=x2y = x^2,
2tx2+x−t−2t3=02tx^2+x-t-2t^3 = 0
 x=tx = t is a root, being PP, so factor it out:
2t(x2−t2)+(x−t)=0⟹(x−t)[2t(x+t)+1]=02t\left(x^2-t^2\right)+\left(x-t\right) = 0 \quad\Longrightarrow\quad \left(x-t\right)\left[2t\left(x+t\right)+1\right] = 0
s=−t−12ts = -t-\frac{1}{2t}
 Both tt and 12t\dfrac1{2t} are positive, so s<0<ts<0<t, and QQ lies on the other side of the axis. Step 3: If a line meets y=x2y = x^2 at x=sx = s and x=tx = t with s<ts<t, the integrand is −(x−s)(x−t)-\left(x-s\right)\left(x-t\right), and its integral over [s,t]\left[s,t\right] is 16(t−s)3\tfrac16\left(t-s\right)^3:
∫st(chord−x2)dx=(t−s)36\int_{s}^{t}\left(\text{chord}-x^2\right)dx = \frac{\left(t-s\right)^3}{6}
 Step 4:
t−s=t−(−t−12t)=2t+12tt-s = t-\left(-t-\frac{1}{2t}\right) = 2t+\frac{1}{2t}
 By AM-GM,
2t+12t ≥ 22t⋅12t=22t+\frac{1}{2t} \ \ge\ 2\sqrt{2t\cdot\frac{1}{2t}} = 2
 with equality when 2t=12t2t = \dfrac{1}{2t}, i.e. t=12t = \dfrac12, which is allowed, since PP is not the origin. Step 5: w↦w3w\mapsto w^3 is increasing for w>0w>0, so
Area=(2t+12t)36 ≥ 236=86=43\text{Area} = \frac{\left(2t+\tfrac{1}{2t}\right)^3}{6} \ \ge\ \frac{2^3}{6} = \frac86 = \frac43
 and the bound is attained:
t=12  ⟹  s=−32,t−s=2,Area=236=43t = \frac12 \implies s = -\frac32,\quad t-s = 2,\quad \text{Area} = \frac{2^3}{6} = \frac43
 Answer: (3) 43\dfrac43
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