Question

If the length of the chord of the parabola $y^2=4x$ whose slope is $1$, is $10\sqrt{2}$ units, then equation of the chord is

• A. $x=y+21$
• B. $4x=4y+21$
• C. $x=y-21$
• D. $4y=4x+21$

Answer 1

The correct option is B $4x=4y+21$

Let the coordinates of the end points of the chord be $(t_1^2,2t_1)$ and $(t_2^2,2t_2)$
$\therefore 1=\frac{2}{t_1+t_2}\Rightarrow t_1+t_2=2\dots \left(1\right)$

Now, the length of the chord is
$10\sqrt{2}=\sqrt{(t_1-t_2{)}^2(4+(t_1+t_2{)}^2)}\Rightarrow 10\sqrt{2}=(t_1-t_2)\sqrt{8}\left[\text{From}\right(1\left)\right]\Rightarrow t_1-t_2=5\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$t_1=\frac{7}{2}$
So, the coordinates of one end of the chord is $(\frac{49}{4},7)$
Equation of the chord is $y-7=x-\frac{49}{4}$
$\therefore 4x=4y+21$