JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
Let $A$, $B$ and $C$ be three points on the parabola $y^{2}=6x$ and let the line segment $AB$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left(\frac{AM\cdot BN}{CD}\right)^{2}$ is equal to $\_\_\_\_\_\_$ .
Parabola
Answer$36$
Solution
Let $A,B,C$ have ordinates $y_1,y_2,y_3$ on $y^2=6x$. $AM=|y_1-y_3|$, $BN=|y_2-y_3|$. Chord $AB$: $6x=(y_1+y_2)y-y_1y_2$. At $y=y_3$: $x_D=\frac{(y_1+y_2)y_3-y_1y_2}{6}$, and $x_C=\frac{y_3^2}6$. $CD=|x_C-x_D|=\frac{|(y_3-y_1)(y_3-y_2)|}{6}$, so $\frac{AM\cdot BN}{CD}=6$, square $=36$.