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(a) \[y+4=0\]

(b) \[x-4=0\]

(c) \[y-4=0\]

(d) \[x+4=0\]

Answer

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Hint: Compare the results of general parabola with given parabola to find “\[a\]”of given parabola.

Given that the focal chord of parabola \[{{y}^{2}}=ax\] is \[2x-y-8=0\].

We know that focal chord is a chord which passes through the focus of parabola.

For standard parabola, \[{{y}^{2}}=4ax\].

Focus is at \[\left( x,y \right)=\left( \dfrac{4a}{4},0 \right)=\left( a,0 \right)\]

Therefore for given parabola, \[{{y}^{2}}=ax\]

We get, focus at \[\left( x,y \right)=\left( \dfrac{a}{4},0 \right)\].

The given focal chord passes through focus.

Therefore, substituting \[x=\dfrac{a}{4},y=0\]in \[2x-y-8=0\]

We get, \[2\left( \dfrac{a}{4} \right)-\left( 0 \right)-8=0\]

\[=\dfrac{a}{2}-8=0\]

Therefore, we get \[a=16\]

Hence, we get parabola \[{{y}^{2}}=ax\]

\[\Rightarrow {{y}^{2}}=16x\]

For general parabola, \[{{y}^{2}}=4ax\]

Directrix is \[x=\dfrac{-4a}{4}\]

\[\Rightarrow x=-a\]

Or \[x+a=0\]

Therefore, for given parabola

\[{{y}^{2}}=16x\]

We get, directrix \[\Rightarrow x=\dfrac{-16}{4}\]

\[\Rightarrow x=-4\]

Or, \[x+4=0\]

Therefore (d) is the correct option.

Note: As we know that, for standard parabola, focus lies on\[x\]axis, we can directly find focus by putting

\[y=0\]in given focal chord which is as follows:

Now, we put \[y=0\]in equation \[2x-y-8=0\].

We get, \[2x-\left( 0 \right)-8=0\]

\[x=\dfrac{8}{2}=4\]

Therefore, focus \[\left( a,0 \right)\]is \[\left( 4,0 \right)\].

Also, a directrix could be found by taking a mirror image of focus through the\[y\]axis which would be

\[\left( -4,0 \right)\].

As we know that the directrix is always perpendicular to the \[x\] axis and passes through \[\left( -4,0 \right)\].

Here, therefore equation of directrix is:

\[x=\text{constant}\]

And here \[\text{constant}=-4\]

Therefore, we get equation of directrix as \[x=-4\]or \[x+4=0\]

Hence, option (d) is correct

Given that the focal chord of parabola \[{{y}^{2}}=ax\] is \[2x-y-8=0\].

We know that focal chord is a chord which passes through the focus of parabola.

For standard parabola, \[{{y}^{2}}=4ax\].

Focus is at \[\left( x,y \right)=\left( \dfrac{4a}{4},0 \right)=\left( a,0 \right)\]

Therefore for given parabola, \[{{y}^{2}}=ax\]

We get, focus at \[\left( x,y \right)=\left( \dfrac{a}{4},0 \right)\].

The given focal chord passes through focus.

Therefore, substituting \[x=\dfrac{a}{4},y=0\]in \[2x-y-8=0\]

We get, \[2\left( \dfrac{a}{4} \right)-\left( 0 \right)-8=0\]

\[=\dfrac{a}{2}-8=0\]

Therefore, we get \[a=16\]

Hence, we get parabola \[{{y}^{2}}=ax\]

\[\Rightarrow {{y}^{2}}=16x\]

For general parabola, \[{{y}^{2}}=4ax\]

Directrix is \[x=\dfrac{-4a}{4}\]

\[\Rightarrow x=-a\]

Or \[x+a=0\]

Therefore, for given parabola

\[{{y}^{2}}=16x\]

We get, directrix \[\Rightarrow x=\dfrac{-16}{4}\]

\[\Rightarrow x=-4\]

Or, \[x+4=0\]

Therefore (d) is the correct option.

Note: As we know that, for standard parabola, focus lies on\[x\]axis, we can directly find focus by putting

\[y=0\]in given focal chord which is as follows:

Now, we put \[y=0\]in equation \[2x-y-8=0\].

We get, \[2x-\left( 0 \right)-8=0\]

\[x=\dfrac{8}{2}=4\]

Therefore, focus \[\left( a,0 \right)\]is \[\left( 4,0 \right)\].

Also, a directrix could be found by taking a mirror image of focus through the\[y\]axis which would be

\[\left( -4,0 \right)\].

As we know that the directrix is always perpendicular to the \[x\] axis and passes through \[\left( -4,0 \right)\].

Here, therefore equation of directrix is:

\[x=\text{constant}\]

And here \[\text{constant}=-4\]

Therefore, we get equation of directrix as \[x=-4\]or \[x+4=0\]

Hence, option (d) is correct