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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Mathematics
Chapter
Parabola

JEE Advanced 2025 · Paper 2 · Mathematics · Parabola

Let S denote the locus of the mid-points of those chords of the...

Let SS denote the locus of the mid-points of those chords of the parabola y2=xy^2=x, such that the area of the region enclosed between the parabola and the chord is 43\frac{4}{3}. Let RR denote the region lying in the first quadrant, enclosed by the parabola y2=xy^2=x, the curve SS, and the lines x=1x=1 and x=4x=4.
Then which of the following statements is (are) TRUE?

Options

  1. A(4,3)∈S(4, \sqrt{3}) \in S
  2. B(5,2)∈S(5, \sqrt{2}) \in S
  3. CArea of RR is 143−23\frac{14}{3}-2 \sqrt{3}
  4. DArea of RR is 143−3\frac{14}{3}-\sqrt{3}

Correct answer

A. (4,3)∈S(4, \sqrt{3}) \in S

Step-by-step solution

T=S1ky−(x+h2)=k2−hx−2ky+2k2−h=0k2−h<0⇒h−k2>0 For  Area : interchange xyy−2kx+2k2−h=0y=x2x2−2kx+(2k2−h)=0<βα∣α−β∣=4k2−4(2k2−h)=4h−4k2A=∫αβ((2kx+h−2k2)−x2)dxA=(4h−4k2)3/26=43(4h−4k2)3/2=8⇒(4h−4k2)=4\begin{aligned} & T=S_1 \\ & k y-\left(\frac{x+h}{2}\right)=k^2-h \\ & x-2 k y+2 k^2-h=0 \\ & k^2-h <0 \quad \Rightarrow h-k^2>0 \\ & \text{\text{ For } Area : interchange } x & y \\ & y-2 k x+2 k^2-h=0 & y=x^2 \\ & x^2-2 k x+\left(2 k^2-h\right)=0 <_\beta^\alpha \\ & |\alpha-\beta|=\sqrt{4 k^2-4\left(2 k^2-h\right)}=\sqrt{4 h-4 k^2} \\ & A=\int_\alpha^\beta\left(\left(2 k x+h-2 k^2\right)-x^2\right) d x \\ & A=\frac{\left(4 h-4 k^2\right)^{3 / 2}}{6}=\frac{4}{3} \\ & \left(4 h-4 k^2\right)^{3 / 2}=8 \quad \Rightarrow\left(4 h-4 k^2\right)=4\end{aligned}\mathrm{h}-\mathrm{k}^2=1 (4, \sqrt{3}) \in \mathrm{S} A=∫14(x−x−1)dx=23(x3/2−(x−1)3/2)14A=23(8−33−1)=23(7−33)=143−23\begin{aligned} & \mathrm{A}=\int_1^4(\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}-1}) \mathrm{dx}=\frac{2}{3}\left(\mathrm{x}^{3 / 2}-(\mathrm{x}-1)^{3 / 2}\right)_1^4 \\ & \mathrm{ A}=\frac{2}{3}(8-3 \sqrt{3}-1)=\frac{2}{3}(7-3 \sqrt{3})=\frac{14}{3}-2 \sqrt{3}\end{aligned}$
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