JEE Advanced 2017 Paper 2 Mathematics: Paragraph 1

With $t_2=-1/t_1$, the length is $2(t_1^2+t_2^2)+4=2\left(t_1+\frac1{t_1}\right)^2$ (using $x=2t^2$ for both ends, the sum of the abscissae plus $4$ simplifies this way). Setting $2(t_1+1/t_1)^2=\frac{25}{2}$ gives $t_1+\frac1{t_1}=\pm\frac52$, so $t_1\in\{2,\tfrac12,-2,-\tfrac12\}$. The slope of the chord joining parameters $t_1,t_2$ is $\dfrac{4(t_1-t_2)}{2(t_1^2-t_2^2)}=\dfrac{2}{t_1+t_2}=\dfrac{2}{t_1-1/t_1}$. For $t_1=2$ this is $\dfrac{2}{3/2}=\dfrac43$, and for $t_1=\tfrac12$ it is $-\dfrac43$. The positive slope is therefore $4/3$. The chord length formula and the slope formula both depend only on the parameter, so one quadratic in t settles everything; the reciprocal pair t and 1/t gives the two symmetric chords of slope plus or minus four thirds.

For parameters $t_1,t_2$ with $t_1t_2=-1$, the midpoint is $M\left(t_1^2+t_2^2,\ 2(t_1+t_2)\right)$. Put $k=t_1+t_2$; then $y=2k$ and $x=t_1^2+t_2^2=k^2-2t_1t_2=k^2+2=\frac{y^2}{4}+2$. So $\mathcal L$ is the parabola $y^2=4(x-2)$, with vertex $(2,0)$. It meets $x=6$ where $y^2=16$, i.e. $y=\pm4$. The enclosed area is $2\displaystyle\int_2^6 2\sqrt{x-2}\,dx=4\cdot\frac23\,(4)^{3/2}=4\cdot\frac23\cdot8=\frac{64}{3}$. Note that the locus is the same parabola translated two units to the right and opened with latus rectum four, which is why the final area uses the segment formula so neatly.