Locus question from Math in focus 3 unit prelim help (1 Viewer)

dabatman · Apr 13, 2017

Please help with the questions, much appreciated

I have attached the questions

1729 · Apr 13, 2017

dabatman said:
Please help with the questions, much appreciated

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. $4a$

dabatman · Apr 13, 2017

1729 said:
$$\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

Locus question from Math in focus 3 unit prelim help (1 Viewer)

dabatman

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