Question

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 1

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

Complete step by step answer:

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 $$(x,y)=({\displaystyle \frac{4a}{4}},0)=(a,0)$$
Therefore for given parabola, $$y^2=ax$$
We get, focus at $$(x,y)=({\displaystyle \frac{a}{4}},0)$$.
The given focal chord passes through focus.
Therefore, substituting $$x={\displaystyle \frac{a}{4}},y=0$$in $$2x-y-8=0$$
We get, $$2\left({\displaystyle \frac{a}{4}}\right)-\left(0\right)-8=0$$
$$={\displaystyle \frac{a}{2}}-8=0$$
Therefore, we get $$a=16$$
Hence, we get parabola $$y^2=ax$$
$$\Rightarrow y^2=16x$$

For general parabola, $$y^2=4ax$$
Directrix is $$x={\displaystyle \frac{-4a}{4}}$$
$$\Rightarrow x=-a$$
Or $$x+a=0$$
Therefore, for given parabola
$$y^2=16x$$
We get, directrix $$\Rightarrow x={\displaystyle \frac{-16}{4}}$$
$$\Rightarrow x=-4$$
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, $$2x-\left(0\right)-8=0$$
$$x={\displaystyle \frac{8}{2}}=4$$
Therefore, focus $$(a,0)$$is $$(4,0)$$.
Also, a directrix could be found by taking a mirror image of focus through the$$y$$axis which would be
 $$(-4,0)$$.
As we know that the directrix is always perpendicular to the $$x$$ axis and passes through $$(-4,0)$$.
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