Question

Find the length of the chord of the parabola ${y^2} = 8x$, whose equation is $x + y = 1$.
A) $8\sqrt 3$
B) $4\sqrt 3$
C) $2\sqrt 3$
D) $\sqrt 3$

Answer 1

Solving the equations of the parabola and its chord, we get the endpoints of the chord. Now using the distance formula we can find the length of the chord. Simplifying we get the answer as an option.

Useful formula:
The second degree equation of the form $a{x^2} + bx + c = 0$ has the solution $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
The distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$

Complete step by step solution:
Given the parabola ${y^2} = 8x$.
And the chord has the equation $x + y = 1$.
To find the length of the chord, first let us find the endpoints of the chord.
For that,
Consider $x + y = 1$
This gives,
$y = 1 - x$
Substituting this in the equation of the parabola we have,
${(1 - x)^2} = 8x$
We have, ${(a - b)^2} = {a^2} - 2ab + {b^2}$
This gives,
${(1 - x)^2} = {1^2} - 2x + {x^2}$
So we have,
${x^2} - 2x + 1 = 8x$
$\Rightarrow {x^2} - 2x + 1 - 8x = 0$
Simplifying we get,
$\Rightarrow {x^2} - 10x + 1 = 0$, which is a second degree equation in $x$.
The second degree equation of the form $a{x^2} + bx + c = 0$ has the solution $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
So we have here
$x = \dfrac{{10 \pm \sqrt {{{( - 10)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{10 \pm \sqrt {100 - 4} }}{2}$
Simplifying we get,
$\Rightarrow x = \dfrac{{10 \pm \sqrt {96} }}{2}$
$\Rightarrow x = 5 \pm 2\sqrt 6$
So we have two cases.
$x = 5 + 2\sqrt 6$ and $x = 5 - 2\sqrt 6$
Substituting in $y = 1 - x$ we get,
$y = - 4 + 2\sqrt 6$ and $y = - 4 - 2\sqrt 6$
So we got the endpoints of the chord as $(5 + 2\sqrt 6 , - 4 + 2\sqrt 6 )$ and $(5 - 2\sqrt 6 , - 4 - 2\sqrt 6 )$
Thus length of the chord can be obtained using the distance formula,
The distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$.
So we get length of the chord as
$l = \sqrt {{{((5 + 2\sqrt 6 ) - (5 - 2\sqrt 6 ))}^2} + {{((5 + 2\sqrt 6 ) - (5 - 2\sqrt 6 ))}^2}}$
$\Rightarrow l = \sqrt {{{(2\sqrt 6 + 2\sqrt 6 )}^2} + (2\sqrt 6 ) + 2\sqrt 6 {)^2}}$
Simplifying we get,
$\Rightarrow l = \sqrt {{{(4\sqrt 6 )}^2} + {{(4\sqrt 6 )}^2}}$
$\Rightarrow l = \sqrt {2 \times 16 \times 6}$
We have, $2 \times 16 \times 6 = 16 \times 12 = {(4 \times 2\sqrt 3 )^2}$
This gives,
Length of the chord equal to $8\sqrt 3$.

Therefore the answer is option A.

Note:
Here we solve for $x$ and substitute to get $y$. Instead we can find $y$ first and substitute to get $x$. Anyway finding the endpoints is important.