JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Ellipse

The directrix $2x+y=6$ has slope $-2$, so the axis of the parabola (perpendicular to the directrix, through the vertex $(2,3)$) has slope $\frac{1}{2}$.

Distance from vertex $(2,3)$ to the directrix: $\frac{|2(2)+3-6|}{\sqrt{5}}=\frac{1}{\sqrt{5}}$; this equals the vertex-to-focus distance $a$.

The foot of the perpendicular from the vertex to the directrix, $Z$, is obtained by moving from the vertex a distance $a$ along the inward normal $\frac{(2,1)}{\sqrt{5}}$ (chosen so as to decrease $2x+y$, since $2(2)+3=7>6$):

$Z=(2,3)-\frac{1}{\sqrt{5}}\cdot \frac{(2,1)}{\sqrt{5}}=(2,3)-\left(\frac{2}{5},\frac{1}{5}\right)=(1.6,2.8)$.

The focus lies on the opposite side of the vertex from $Z$, at the same distance, so the vertex is the midpoint of focus and $Z$:

$\text{Focus}=2(2,3)-(1.6,2.8)=(2.4,3.2)$.

The ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (with $a>b$) has eccentricity $\frac{1}{\sqrt{2}}$, so $b^2=a^2(1-e^2)=\frac{a^2}{2}$, i.e. $a^2=2b^2$.

Since $E$ passes through the focus $(2.4,3.2)$:

$\frac{(2.4{)}^2}{2b^2}+\frac{(3.2{)}^2}{b^2}=1$

$\frac{5.76}{2b^2}+\frac{10.24}{b^2}=1$

$\frac{2.88+10.24}{b^2}=1\Rightarrow b^2=13.12=\frac{328}{25}$.

The square of the latus rectum length is ${\left(\frac{2b^2}{a}\right)}^2=\frac{4b^4}{a^2}=\frac{4b^4}{2b^2}=2b^2=2\cdot \frac{328}{25}=\frac{656}{25}$.