JEE Main 2019 (Online) 10th January Evening ShiftQuestion 1
The length of the chord of the parabola $x^2=4y$ having equation $x-\sqrt2y+4\sqrt2=0$ is :
Parabola
Solution
Concept. Solve the line with the parabola for $x$; the chord length is $|x_1-x_2|\sqrt{1+m^2}$. Formulas used. (1) $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$.(2) Chord on $y=mx+c$: length $=|x_1-x_2|\sqrt{1+m^2}$.Step 1 - The line. $\sqrt2y=x+4\sqrt2\Rightarrow y=\frac{x}{\sqrt2}+4$, so $m=\frac1{\sqrt2}$. Step 2 - Meet $x^2=4y$. $x^2=\frac{4x}{\sqrt2}+16\Rightarrow x^2-2\sqrt2x-16=0$. Step 3 - Use (1). $(x_1-x_2)^2=8+64=72\Rightarrow|x_1-x_2|=6\sqrt2$. (The ends are $A(4\sqrt2,8)$ and $B(-2\sqrt2,2)$.)Step 4 - Use (2). $6\sqrt2\cdot\sqrt{1+\frac12}=6\sqrt2\cdot\frac{\sqrt3}{\sqrt2}=6\sqrt3$. Answer: A