If the x-intercept of a focal chord of the parabola y 2 = 8 x + 4 y + 4 is 3, then the length of this chord is equal to _______. [2023]

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Solution

Q.
If the x-intercept of a focal chord of the parabola $y^{2} = 8 x + 4 y + 4$ is 3, then the length of this chord is equal to _______. [2023]

Ans.
(16)

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

$y^{2} = 8 x + 4 y + 4 \Rightarrow {(y - 2)}^{2} = 8 (x + 1) \Rightarrow Y^{2} = 4 \cdot 2 \cdot X$

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

For focus $x + 1 = 2$ and $y - 2 = 0 \Rightarrow x = 1$ and $y = 2$

$∴$ $Focus is (1, 2)$

Equation of line passing through $(1, 2)$ is

$y - 2 = m (x - 1),$ $(3, 0)$ lies on the above line,

$∴$    $- 2 = m (3 - 1) \Rightarrow m = - 1$

$∴$   Equation of line is $y - 2 = - 1 (x - 1)$

i.e., $x + y - 3 = 0$ $∴$ $y = 3 - x$

Now put the value of $y$ in equation of parabola ${(3 - x)}^{2} = 8 x + 4 (3 - x) + 4$

$\Rightarrow 9 + x^{2} - 6 x = 8 x + 12 - 4 x + 4$    $∴$    $x = 5 + 4 \sqrt{2}, 5 - 4 \sqrt{2}$

So, $y = - 2 - 4 \sqrt{2},$    $4 \sqrt{2} - 2$

So, length of focal chord is

$= \sqrt{{(4 \sqrt{2} + 4 \sqrt{2})}^{2} + {(- 2 - 4 \sqrt{2} - 4 \sqrt{2} + 2)}^{2}}$

$= \sqrt{128 + 128} = \sqrt{256} = 16$

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