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Q.
A chord is drawn through the focus of the parabola $y^2 = 6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt{5}}{2}$, then its slope can be :-
Equation of parabola, $y^{2} = 6x$
$\Rightarrow v^{2} = 4 \times\frac{3}{2}x$
$\therefore $ Focus $= \left(\frac{3}{2}, 0\right)$
Let equation of chord passing through focus be
$ax + by+ c =0 \,\ldots\left(1\right)$
Since chord is passing through $\left(\frac{3}{2}, 0\right)$
$\therefore $ Put $x = \frac{3}{2}, y = 0$ in eqn $\left(1\right)$, we get
$\frac{3}{2}a + c = 0$
$\Rightarrow c = -\frac{3}{2}a\quad...\left(2\right)$
distance of chord from origin is $\frac{\sqrt{5}}{2}$
$\frac{\sqrt{5}}{2} = \left|\frac{a\left(0\right)+b\left(0\right)+c}{\sqrt{a^{2}+b^{2}}}\right| = \frac{c}{\sqrt{a^{2}+b^{2}}}$
Squaring both sides
$\frac{5}{4} = \frac{c^{2}}{a^{2}+b^{2}}$
$\Rightarrow a^{2}+b^{2} = \frac{4}{5}c^{2}$
Putting value of c from $\left(2\right)$, we get
$a^{2}+b^{2} = \frac{4}{5}\times\frac{9}{4}a^{2}$
$b^{2} = \frac{9}{4}a^{2}-a^{2} = \frac{4}{5}a^{2}$
$\frac{a^{2}}{b^{2}} = \frac{5}{4}, \frac{a}{b} = \pm\frac{\sqrt{5}}{2}$
Slope of chord, $\frac{dy}{dx}$
$= -\frac{a}{b} = -\left(\frac{\pm\sqrt{5}}{2}\right) = \mp$
$ \frac{\sqrt{5}}{2}$