# Point M(4,-8) lies on `x^2=-2y` .a. Find the equation of the focal chord through M.b. Find point N where this chord cuts the parabola again.

lemjay | High School Teacher | (Level 2) Senior Educator

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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)=-15/8(x-0)`

`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=-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`

`(x-4)(4x+1)=0`

Set each factor to zero and solve for x.

`x-4=0`               and               `4x+1=0`

`x=4`                                          `x=-1/4`

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)` .