Question

If the $x$-intercept of the focal chord of the parabola $y^2-2y-4x+5=0$ is $\frac{11}{4}$, then find the length of chord.

• A. $\frac{15}{2}$
• B. $\frac{25}{2}$
• C. $\frac{25}{4}$
• D. $\frac{15}{4}$

Answer 1

The correct option is C $\frac{25}{4}$

$y^2-2y-4x+5=0\Rightarrow y^2-2y+1=4x-4\Rightarrow (y-1{)}^2=4(x-1)$

Vertex $:(1,1)$, Focus $:(2,1)$
Focal chord passes through $(\frac{11}{4},0)$
Therefore, equation of focal chord is
$\frac{y}{x-\frac{11}{4}}=\frac{1}{2-\frac{11}{4}}$

$\Rightarrow 4x+3y=11$
Let a point on parabola, through which focal chord passes be $(1+t^2,1+2t)$
$\because 4x+3y=11\Rightarrow 4(1+t^2)+3(1+2t)=11\Rightarrow (2t-1)(t+2)=0\Rightarrow t=-2,\frac{1}{2}$

Therefore, the end points of the focal chord are :
$(1+t_1^2,1+2t_1),(1+t_2^2,1+2t_2)$

Length of chord, $L=\sqrt{(t_2^2-t_1^2{)}^2+(2t_2-2t_1{)}^2}$
$\Rightarrow L=\sqrt{{(4-\frac{1}{4})}^2+(-4-1{)}^2}=\frac{25}{4}$