# Thread: Locus question from Math in focus 3 unit prelim help

1. ## Locus question from Math in focus 3 unit prelim help

I have attached the questions

2. ## Re: Locus question from Math in focus 3 unit prelim help

Originally Posted by dabatman

I have attached the questions
$\noindent \textbf{Q9.} The parabola x^2 = -6y has focus S\left(0, -\frac{3}{2}\right). The focal chord is simply a chord of the parabola which passes through the focal point, and so the chord passes through (6, - 6) and S\left(0, -\frac{3}{2}\right). From here, the equation of the focal chord can be found along with the coordinates of X through simultaneous equations.$

$\noindent \textbf{Q14.} The point Q(2aq, aq^2) lies on the parabola x^2 = 4ay, as can be shown through substitution. The focus has coordinates S(0, a) and so the gradient of the required equation is given by m = \frac{a - aq^2}{-2aq} = \frac{q^2 - 1}{2q}. Thus, the required equation y = mx + a is y = \left(\frac{q^2 - 1}{2q}\right)x + a \Rightarrow (q^2 - 1)x - 2qy + 2aq = 0. \\\\ The latus rectum is a form of this equation where q = 1. That is, y = a. Now let q = 1 in Q(2aq, aq^2) to obtain the endpoints A(2a, a), and through symmetry, B(-2a, a). The length of the latus rectum is given by the length of the interval AB, i.e. 4a$

3. ## Re: Locus question from Math in focus 3 unit prelim help

Originally Posted by 1729
$\noindent \textbf{Q9.} The parabola x^2 = -6y has focus S\left(0, -\frac{3}{2}\right). The focal chord is simply a chord of the parabola which passes through the focal point, and so the chord passes through (6, - 6) and S\left(0, -\frac{3}{2}\right). From here, the equation of the focal chord can be found along with the coordinates of X through simultaneous equations.$

$\noindent \textbf{Q14.} The point Q(2aq, aq^2) lies on the parabola x^2 = 4ay, as can be shown through substitution. The focus has coordinates S(0, a) and so the gradient of the required equation is given by m = \frac{a - aq^2}{-2aq} = \frac{q^2 - 1}{2q}. Thus, the required equation y = mx + a is y = \left(\frac{q^2 - 1}{2q}\right)x + a \Rightarrow (q^2 - 1)x - 2qy + 2aq = 0. \\\\ The latus rectum is a form of this equation where q = 1. That is, y = a. Now let q = 1 in Q(2aq, aq^2) to obtain the endpoints A(2a, a), and through symmetry, B(-2a, a). The length of the latus rectum is given by the length of the interval AB, i.e. 4a$
thanks so much

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