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
Verified
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
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