JEE Main 2025 Jan 23 Shift 2, Mathematics Q15: Ellipse: Chord, Tangent and Normal

For a chord with given midpoint $(x_0,y_0)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the chord equation is $T=S_1$:

$\frac{x\cdot 1}{4}+\frac{y\cdot \frac{1}{2}}{2}=\frac{1}{4}+\frac{1/4}{2}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}.$

Multiply by $4$: $x+y=\frac{3}{2}$. Substitute $y=\frac{3}{2}-x$ into the ellipse:

$x^2+2{\left(\frac{3}{2}-x\right)}^2=4\Rightarrow x^2+2\left(\frac{9}{4}-3x+x^2\right)=4\Rightarrow 3x^2-6x+\frac{1}{2}=0,$

or $6x^2-12x+1=0$. Roots $x_1,x_2$ satisfy $x_1+x_2=2$, $x_1{x}_2=\frac{1}{6}$.

$(x_2-x_1{)}^2=(x_1+x_2{)}^2-4x_1{x}_2=4-\frac{4}{6}=\frac{10}{3}.$

Since the chord has slope $-1$ (from $x+y=\frac{3}{2}$), $|y_2-y_1|=|x_2-x_1|$, so the chord length is $\sqrt{2}|x_2-x_1|=\sqrt{2}\cdot \sqrt{\frac{10}{3}}=\sqrt{\frac{20}{3}}=\frac{2\sqrt{15}}{3}.$