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A.\[\dfrac{{320}}{7}\]

B.\[\dfrac{{320}}{9}\]

C.\[\dfrac{{80}}{9}\]

D.\[\dfrac{{640}}{7}\]

Answer

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We have given the equation of the chord \[4y = 3x + 8\] of the parabola\[{y^2} = 8x\]. We have to find the length of the chord.

Consider the given equation of chord:

\[4y = 3x + 8\]

Now, we solve the equation for\[x\]:

\[4y - 8 = 3x\]

\[x = \dfrac{{4y - 8}}{3}\] … (1)

Consider the given equation of the parabola:

\[{y^2} = 8x\]

Now, we substitute \[x = \dfrac{{4y - 8}}{3}\] into the above equation:

$\Rightarrow$ \[{y^2} = 8\left( {\dfrac{{4y - 8}}{3}} \right)\]

$\Rightarrow$ \[3{y^2} = 32y - 64\]

$\Rightarrow$ \[3{y^2} - 32y + 64 = 0\]

Factorize the expression by splitting the middle term.

\[3{y^2} - 24y - 8y + 64 = 0\]

$\Rightarrow$ \[3y\left( {y - 8} \right) - 8\left( {y - 8} \right) = 0\]

$\Rightarrow$ \[\left( {y - 8} \right)\left( {3y - 8} \right) = 0\]

Then, the obtained values of \[y\]are:

$\Rightarrow$ \[y = 8{\rm{ or }} y = \dfrac{8}{3}\]

Now, we substitute the value of \[y\] into the equation (1) and evaluate the corresponding values of\[x\].

\[x = \dfrac{{4\left( 8 \right) - 8}}{3}\]

$\Rightarrow$ \[x = \dfrac{{24}}{3}\]

$\Rightarrow$ \[x = 8\]

Similarly, \[x = \dfrac{{4\left( {\dfrac{8}{3}} \right) - 8}}{3}\]

\[x = \dfrac{{\dfrac{{32}}{3} - 8}}{3}\]

$\Rightarrow$ \[x = \dfrac{{32 - 24}}{9}\]

$\Rightarrow$ \[x = \dfrac{8}{9}\]

The obtained coordinates are \[\left( {8,8} \right)\] and \[\left( {\dfrac{8}{9},\dfrac{8}{3}} \right)\].

These coordinates are the end point of the chord. Now, we use the formula of distance between two points to find the length of the chord.

Assume \[\left( {{x_1},{y_1}} \right) = \left( {8,8} \right)\] and \[\left( {{x_2},{y_2}} \right) = \left( {\dfrac{8}{9},\dfrac{8}{3}} \right)\] , then we know that the distance between the points \[\left( {{x_1},{y_1}} \right)\]and\[\left( {{x_2},{y_2}} \right)\]is given as:

\[L = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Substitute \[\left( {{x_1},{y_1}} \right) = \left( {8,8} \right)\] and \[\left( {{x_2},{y_2}} \right) = \left( {\dfrac{8}{9},\dfrac{8}{3}} \right)\] into the above formula:

$\Rightarrow$ \[L = \sqrt {{{\left( {\dfrac{8}{9} - 8} \right)}^2} + {{\left( {\dfrac{8}{3} - 8} \right)}^2}} \]

Evaluate the chord length:

\[L = \sqrt {{{\left( { - \dfrac{{64}}{9}} \right)}^2} + {{\left( { - \dfrac{{16}}{3}} \right)}^2}} \]

$\Rightarrow$ \[L = \sqrt {\dfrac{{4069}}{{81}} - \dfrac{{256}}{9}} \]

$\Rightarrow$ \[L = \sqrt {\dfrac{{4096 + 2304}}{{81}}} \]

$\Rightarrow$ \[L = \sqrt {\dfrac{{6400}}{{81}}} \]

$\Rightarrow$ \[L = \dfrac{{80}}{9}\]

Therefore, the length of the chord is \[\dfrac{{80}}{9}\] units.

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