Parabola question (1 Viewer)

[ratpack90]

Ahhh.. need help with the second half of the working out of this question

1. The normal of the parabola x^2=18y at (-6,2) cuts the paraola again at Q. Find Q.

Oh and also

2. The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates of X.

iI really appreciate anyone's input

 

[zingerburger]

For both questions, normally you'd have to find the equation or gradient of the tangent first. If you've found the gradient, you can find the equation of the line using y-y1 = m(x-x1) and substituting y1 and x1 with the coordinates of the question. Then rearrange the equation to get y =.

Also, rearrange the equation of the parabolas to make y the subject. Once you have both equations of the line and the parabola, it's a simple matter of solving them simultaneously. Since both equations have y as the subject, you can equate them, get all x's on one side to make the equation equal zero, and you'll have a quadratic to solve for x.

You'll get two values of x - one being the value that's already in the question and one that you'll use to substitute into either the line or parabola equation to find the value of y. This will finally give you the coordinates of intersection.

I hope that helps.

 

[ratpack90]

Ahh..awesome yeh thanks for that it helped a lot

 

[ratpack90]

well your advice worked for my first question but im stuck on the second question involving the focal chord

 

[SSejychan]

2. The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates of X.

Because the parabola is concave down, the general equation of the parabola is x^2 = -4ay. Now, equate this with the equation of the parabola x^2 = -6y, to find the value of a.

Now, you have to find the vertex of the parabola. The vertex of the parabola is at (0,0) because the equation is not (x-h)^2 = -4a(y-k).

Because the parabola is concave down, thus the focus will have coordinates (0,-a)

now, using the focus (0,-a) and one of the points where the focal chord cuts the parabola (6,-6), you can find the gradient of the focal chord, and thus use y - y1 = m(x - x1) to find the equation of the focal chord.

Now, using this equation you can solve it (like zingerburger said) by rearranging the equations (focal chord and parabola) to get y as the subject, then solving simultaneously. Then you'll have a quadratic which will give you two values of x --> one which was already given to you, and one which you were supposed to find. Then, you can find the y-coordinate for this x-value, and now you have the co-ordinates of 'X'.

 

[ratpack90]

Cool thanks yeh managed to get the right answer. Like i said before i appreciate everyone's input.

Oh and i love the FMA pic

 

[SSejychan]

Cool thanks yeh managed to get the right answer. Like i said before i appreciate everyone's input.

Oh and i love the FMA pic

only glad to help.

<3 fma.