1 Answer | Add Yours
In parabola, when it is the variable x that has an exponent 2, then the parabola either opens upward or downward and its axis of symmetry is vertical. Its standard equation is:
`(x-h)^2 = 4a(y-k)`
where (h,k) - is the vertex of the parabola and
a - is the distance between the vertex & focus, as well as the distance between vertex & directrix.
Also, if 4a is positive, the parabola opens upward. And if negative, the parabola opens downward.
For the given equation `x^2 = -2y` , its vertex is (0,0) ans since 4a is negative, the parabola opens downward.
Then let's solve for the focus. To do so, determine value of a.
`4a = -2` `a = -2/4` `a=-1/2`
Note that focus lie on the axis of symmetry of the parabola. And since the axis of symmetry is vertical, the focus is:
`F(x,y+a) = F(0 , 0 + (-1/2)) = F(0, 0-1/2) = F(0,-1/2)`
(a) To solve for the equation of the chord that contains F(0,-1/2) and M(4,-8), let's determine the slope of the line first.
`m = (y_2 - y_1)/(x_2 - x_1) = (-8 - (-1/2))/(4-0) = (-15/2)/4 = -15/8`
Then, use the point-slope formula of a line.
`y-y_1 = m(x-x_1)`
`y+1/2 = -15/8x`
`y= -15/8x - 1/2`
Hence, the equation of the focal chord that passes through the point M(4,-8) is `y=-15/8x - 1/2` .
(b)To determine point N, let's solve for the intersection points between the parabola and the equation of the line.
To do so, let's use substitution method of system of equations.
Substitute `y=-15/8x-1/2` to `x^2=-2y` .
`x^2=-2(-15/8x - 1/2)`
`x^2 = 15/4x +1`
Then, express the equation in quadratic form `ax^2+bx+c=0` .
`x^2-15/4x - 1 =0`
`4x^2 - 15x - 4 =0`
Set each factor to zero and solve for x.
`x-4=0` and `4x+1=0`
Then, substitute values of x to the equation of the line.
`x=4` , `y=-15/8(4)-1/2 = -15/2-1/2= -16/2=-8`
`x=-1/4` , `y=-15/8(-1/4)-1/2=15/32-1/2=15/32-16/32=-1/32`
So, the parabola `x^2=-2y` and the focal chord `y=-15/8x-1/2` intersects at (4,8) and (-1/4,-1/32).
Hence the focal chord `y=-15/8x-1/2` cuts the parabola again at `N(-1/4,-1/32)` .
We’ve answered 317,416 questions. We can answer yours, too.Ask a question