Question 6

**Setting up:** Points on $y^2=x$: $(t_1^2,t_1),(t_2^2,t_2)$. Area between chord and parabola $=\dfrac{(t_2-t_1)^3}{6}=\dfrac43 \Rightarrow t_2-t_1=2$. Midpoint $(h,k)$: $k=\dfrac{t_1+t_2}{2}$, $h=\dfrac{t_1^2+t_2^2}{2}=\dfrac{(t_1+t_2)^2+(t_2-t_1)^2}{4}=k^2+1$. So $S: y^2=x-1$. - (A) $(4,\sqrt3)$: $3=4-1$ ✓ - (B) $(5,\sqrt2)$: $2\ne 4$ ✗ **Area of $\mathcal R$:** $\displaystyle\int_1^4(\sqrt{x}-\sqrt{x-1})\,dx=\frac23(8-1)-\frac23(3\sqrt3)=\frac{14}{3}-2\sqrt3$. So (C) is true.

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