If the x -intercept of a focal chord of the parabola y ^ 2 = 8x +...

Given,

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

Now simplifying equation of the given parabola we get,

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

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

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

$a = 2, X = x + 1, Y = y - 2$

Hence, the vertex is \left(- 1, 2\right)$and focus is \left(1, 2\right)$

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

Now, putting \left(3, 0\right)${as given x -intercept of focal chord is 3} in the above line we get,$m = - 1$</p><p>So, equation of chord will be$y = 3 - x$,</p><p>Now putting the value of$y = 3 - x$in parabola${\left(y - 2\right)}^{2} = 8\left(x + 1\right) we get,

$⇒ {\left(1 - x\right)}^{2} = 8\left(x + 1\right)$

$⇒ {x}^{2} - 10x - 7 = 0$

$⇒ \left|{x}_{1} - {x}_{2}\right| = \sqrt{{10}^{2} + 4 \times 7} = \sqrt{128}$

$⇒ \left|{y}_{1} - {y}_{2}\right| = \left|3 - {x}_{1} - \left(3 - {x}_{2}\right)\right| = \left|{x}_{2} - {x}_{1}\right| = \sqrt{128}$

Length of focal chord $\sqrt{{\left|{x}_{2} - {x}_{1}\right|}^{2} + {\left|{y}_{2} - {y}_{1}\right|}^{2}} = \sqrt{256} = 16$.