If the x-intercept of a focal chord of the parabola y^2=8x+4y+4 is 3, then the length of this chord is equal to .

Given,

The $x$x-intercept of a focal chord of the parabola ${y}^{2}=8x+4y+4$y2=8x+4y+4 is $3$3,

Now simplifying equation of the given parabola we get,

$${y}^{2}=8x+4y+4$$y2=8x+4y+4

$$\Rightarrow {(y-2)}^{2}=8(x+1)$$⇒(y−2)2=8(x+1)

Now on comparing with standard parabola ${y}^{2}=4ax$y2=4ax we get,

$$a=2,X=x+1,Y=y-2$$a=2,X=x+1,Y=y−2

Hence, the vertex is $(-1,2)$(−1,2) and focus is $(1,2)$(1,2)

So, equation of line passing through focus is given by $y-2=m(x-1)$y−2=m(x−1)

Now, putting $(3,0)$(3,0) {as given $x-$x−intercept of focal chord is $3$3} in the above line we get, $m=-1$m=−1

So, equation of chord will be $y=3-x$y=3−x,

Now putting the value of $y=3-x$y=3−x in parabola ${(y-2)}^{2}=8(x+1)$(y−2)2=8(x+1) we get,

$$\Rightarrow {(1-x)}^{2}=8(x+1)$$⇒(1−x)2=8(x+1)

$$\Rightarrow {x}^{2}-10x-7=0$$⇒x2−10x−7=0

$$\Rightarrow |{x}_{1}-{x}_{2}|=\sqrt{{10}^{2}+4\times 7}=\sqrt{128}$$⇒∣x1​−x2​∣=102+4×7​=128​

$$\Rightarrow |{y}_{1}-{y}_{2}|=|3-{x}_{1}-(3-{x}_{2})|=|{x}_{2}-{x}_{1}|=\sqrt{128}$$⇒∣y1​−y2​∣=∣3−x1​−(3−x2​)∣=∣x2​−x1​∣=128​

Length of focal chord $\sqrt{{|{x}_{2}-{x}_{1}|}^{2}+{|{y}_{2}-{y}_{1}|}^{2}}=\sqrt{256}=16$∣x2​−x1​∣2+∣y2​−y1​∣2​=256​=16.