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chord is drawn through the focus of the parabola
$y^{2} = 6 x$ such that its distance from the vertex of
this parabola is $\frac{\sqrt{5}}{2}$ then its slope can be:

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detailed solution

Correct option is A

Let the equation of the chord be.y−0=mx−32⇒y=mx-3m2 Its distance from the vertex (0, 0) is ca2+b2= m−321+m2=523m1+m2=5squaring on both sides9m2=51+m2 4m2=5 m=±52.

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chord is drawn through the focus of the parabolay2=6x such that its distance from the vertex ofthis parabola is 52then its slope can be:

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Ready to Test Your Skills?

free study material

chord is drawn through the focus of the parabola
$y^{2} = 6 x$ such that its distance from the vertex of
this parabola is $\frac{\sqrt{5}}{2}$ then its slope can be: