Let S denote the locus of the mid-points of those chords of the parabola y^2=x , such that the area of the region enclosed between the parabola and the chord is \frac43 . Let \mathcalR denote the region lying in the first quadrant, enclosed by the parabola y^2=x , the curve S , and the lines x=1 and x=4 . Then which of the following statements is (are) TRUE?

(4, \sqrt3) \in S, Area of \mathcalR is \frac143 - 2\sqrt3

Area of \mathcalR is \frac143 - 2\sqrt3

Area of \mathcalR is \frac143 - \sqrt3

Parabola PYQ — JEE Advanced 2025 · Shift 2 · Q22 with Solution

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Correct answer: A, C

For the parabola $y^2=x$ , a standard parametric point is $P(t)=(t^2,t).$ Take a chord joining $P(t_1)=(t_1^2,t_1), \qquad P(t_2)=(t_2^2,t_2).$

Its midpoint is $M\left(\frac{t_1^2+t_2^2}{2},\frac{t_1+t_2}{2}\right).$

Write the parabola as $x=y^2$ . Since the chord joins $P(t_1)$ and $P(t_2)$ , its equation in the form $x=\text{linear in }y$ is obtained as follows.

Slope in the $(x,y)$ -plane is not the most convenient; instead use the two-point form with $x$ as a function of $y$ : $x=(t_1+t_2)y-t_1t_2.$

Hence the enclosed area is $A=\int_{t_1}^{t_2}\big[(t_1+t_2)y-t_1t_2-y^2\big]dy.$

Let $s=t_1+t_2, \qquad p=t_1t_2.$ Then $A=\int_{t_1}^{t_2}(sy-p-y^2)dy.$

Now use the standard result (or simplify directly): $A=\frac{(t_2-t_1)^3}{6}.$

Since the given area is $\frac43$ , $\frac{(t_2-t_1)^3}{6}=\frac43$ $\Rightarrow (t_2-t_1)^3=8$ $\Rightarrow t_2-t_1=2.$ (We take $t_2>t_1$ .)

Let the midpoint be $(h,k)$ . Then $k=\frac{t_1+t_2}{2}=\frac{s}{2}.$ Also, $h=\frac{t_1^2+t_2^2}{2}= rac{(t_1+t_2)^2-2t_1t_2}{2}.$

Using $(t_2-t_1)^2=(t_1+t_2)^2-4t_1t_2,$ and $t_2-t_1=2$ , we get $4=s^2-4p \Rightarrow p=\frac{s^2-4}{4}.$

Therefore,

=\frac{s^2-2\cdot \frac{s^2-4}{4}}{2} =\frac{s^2+4}{4}.$$ Since $s=2k$, $$h=\frac{4k^2+4}{4}=k^2+1.$$ So the locus $S$ is $$x=y^2+1.$$ In the first quadrant, this is $$y=\sqrt{x-1}.$$ --- 4. **Check options A and B** Since points on $S$ satisfy $x=y^2+1$: - For $(4,\sqrt3)$: $$y^2+1=3+1=4,$$ so $(4,\sqrt3)\in S$. Hence **A is true**. - For $(5,\sqrt2)$: $$y^2+1=2+1=3\ne 5,$$ so $(5,\sqrt2)\notin S$. Hence **B is false**. --- 5. **Region $\mathcal R$ in the first quadrant** The parabola is $$y=\sqrt{x},$$ and the locus $S$ is $$y=\sqrt{x-1}.$$ Between $x=1$ and $x=4$, the parabola lies above $S$ because $$\sqrt{x}>\sqrt{x-1}.$$ Therefore area of $\mathcal R$ is $$\int_1^4 \left(\sqrt{x}-\sqrt{x-1}\right)dx.$$ Compute separately: $$\int_1^4 \sqrt{x}\,dx=\left[\frac{2}{3}x^{3/2}\right]_1^4 =\frac{2}{3}(8-1)=\frac{14}{3}.$$ Also, $$\int_1^4 \sqrt{x-1}\,dx=\int_0^3 \sqrt{u}\,du =\left[\frac{2}{3}u^{3/2}\right]_0^3 =\frac{2}{3}(3\sqrt3)=2\sqrt3.$$ Hence, $$\text{Area}(\mathcal R)=\frac{14}{3}-2\sqrt3.$$ So **C is true** and **D is false**. --- 6. **Final conclusion** The true statements are: $$\boxed{A \text{ and } C}$$ This matches the stored correct answer.