JEE Main 2025 Apr 4 Shift 1, Mathematics Q24: Ellipse: Chord, Tangent and Normal

Locating $P$. The tangent to the circle $x^2+(y-1{)}^2=2$ at $(x_1,y_1)$ is $x{x}_1+(y-1)(y_1-1)=2$, which on expanding becomes $x{x}_1+y(y_1-1)=y_1+1$.
Matching with $x+y=3$: $x_1=1$, $y_1-1=1$ (so $y_1=2$), and $y_1+1=3$ also give $y_1=2$. So $P=(1,2)$.

Locating $Q$ and $R$. Parametrise the line $x+y=3$ from $P$ along the unit direction $\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$ (or its negative). Points at distance $\frac{2\sqrt{2}}{3}$ from $P$:

$\left(1\pm \frac{2}{3},2\mp \frac{2}{3}\right)=\left(\frac{5}{3},\frac{4}{3}\right)\text{ or }\left(\frac{1}{3},\frac{8}{3}\right)$.

Since $E_1$ has major axis along $x$-axis (wider in $x$) and $E_2$ has major axis along $y$-axis, by symmetry the contact with $E_1$ has the larger $x$-coordinate: $Q=\left(\frac{5}{3},\frac{4}{3}\right)$ and $R=\left(\frac{1}{3},\frac{8}{3}\right)$. Note that $P$ is indeed the midpoint of $QR$.

Final value.

$x_1{y}_1+x_2{y}_2+x_3{y}_3=1\cdot 2+\frac{5}{3}\cdot \frac{4}{3}+\frac{1}{3}\cdot \frac{8}{3}=2+\frac{20}{9}+\frac{8}{9}=2+\frac{28}{9}=\frac{46}{9}$.

So $9\times \frac{46}{9}=46$.