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If a focal chord of the parabola \[{{y}^{2}}=ax\]is \[2x-y-8=0\], then the equation of the directrix is:
(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