A chord is drawn through the focus of the parabola y^2=6x such that its distance from the …

Question

A chord is drawn through the focus of the parabola $y^2=6x$ such that its distance from the vertex of this parabola is $\dfrac{\sqrt5}{2}$; then its slope can be:

Accepted Answer

Here $4a=6\Rightarrow a=	frac32$, focus $=\left(	frac32,0ight)$. Let the chord be $\ell x+my+n=0$ through the focus, so $n=-	frac32\ell$.

Distance from vertex $(0,0)$: $\dfrac{|n|}{\sqrt{\ell^2+m^2}}=\dfrac{\sqrt5}{2}\Rightarrow \dfrac{	frac94\ell^2}{\ell^2+m^2}=\dfrac54\Rightarrow \dfrac{\ell^2}{m^2}=\dfrac54.$

Slope $=-\dfrac{\ell}{m}=\mp\dfrac{\sqrt5}{2}$, so the slope can be $\dfrac{\sqrt5}{2}$ (in magnitude).

Solutions · 1