Answer any five (5) Questions.

Question

ScanSolve AI Tutor · Accepted Answer

Okay, cloths, let's find where the locus meets the y-axis.

Set $x=0$ in the equation of the locus.
The equation of the locus is $3x^2 + 3y^2 + 12x - 38y + 55 = 0$ .
Substitute $x=0$ :
$3(0)^2 + 3y^2 + 12(0) - 38y + 55 = 0$ $0 + 3y^2 + 0 - 38y + 55 = 0$ $3y^2 - 38y + 55 = 0$
Solve the quadratic equation for $y$ .
We can use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=3$ , $b=-38$ , and $c=55$ .
$y = \frac{-(-38) \pm \sqrt{(-38)^2 - 4(3)(55)}}{2(3)}$ $y = \frac{38 \pm \sqrt{1444 - 660}}{6}$ $y = \frac{38 \pm \sqrt{784}}{6}$ $y = \frac{38 \pm 28}{6}$
Calculate the two possible values for $y$ .
$y_1 = \frac{38 + 28}{6} = \frac{66}{6} = 11$ $y_2 = \frac{38 - 28}{6} = \frac{10}{6} = \frac{5}{3}$
State the points where the locus meets the y-axis.
Since $x=0$ for both points, the points are $(0, y_1)$ and $(0, y_2)$ .
The points are $(0, 11) \text{ and } \left(0, \frac{5}{3}\right)$ .

Answer