Question

The distance between the vertex and the focus of parabola x^2-4x+3y-8 is

Answer 1

Answer:

3/4

Step-by-step explanation:

Hello There...

From the given equation of parabola$x^{2}-4x+3y-8=0$

We are going to make it into a familiar form

$(x-a)^{2}=4p(y-b)$..........(1)

(The standard equation of parabola)

$3y=-x^{2}+4x+8$

$=3y=-(x^{2} -4x-8)$

$=y=-\frac{1}{3}(x-2)^{2}+4$

$=y-4=-\frac{1}{3}(x-2)^{2}$

$=3(y-4)=(x-2)^{2}$

$=-4*\frac{3}{4}(y-4)=(x-2)^{2}$

In eq (1)

p is the distance between focus and vertex

Here p=-3/4

=>Distance is 3/4 from left side of the vertex(-ve sign => left side of vertex since the parabola is opened on the left side)

Hope this would help

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