In the diagram below, O (0 ; 0) is the centre of circle ABC with A(-4; -3) and C(4; -3) - NSC Technical Mathematics - Question 2 - 2019 - Paper 2

To find the coordinates of point B where the tangent touches the circle, we use the fact that B lies on the circle's circumference. Since information about B isn't provided directly, we can use the fact that the tangent PQ is parallel to MN and lies at the same y-coordinate as the center O. Therefore, knowing the center: $B(0; 5)$

The equation of tangent PQ is needed to find its gradient. The tangent MN has a gradient of -\frac{4}{3} since its equation is given as y = -\frac{4}{3}x + \frac{25}{3}. Since tangents are parallel, Thus, the gradient of PQ is also: $m_{PQ} = -\frac{4}{3}$

Using the gradient found and point B(0, 5), we can write the equation of line PQ: Using point-slope form: $y - y_1 = m(x - x_1)$ We have: $y - 5 = -\frac{4}{3}(x - 0)$ Rearranging gives: $y = -\frac{4}{3}x + 5$

Starting with the given equation: $x^2 + 8y^2 - 32 = 0$ Rearranging gives: $x^2 + 8y^2 = 32$ Dividing both sides by 32 gives: $\frac{x^2}{32} + \frac{y^2}{4} = 1$ This means: $a^2 = 32, b^2 = 4$

The graph defined by the equation presents as an ellipse. The x-intercepts occur where y = 0: Substituting gives: $\frac{x^2}{32} = 1 \Rightarrow x = \pm 4\sqrt{2}$ The y-intercepts occur where x = 0: Substituting provides: $\frac{y^2}{4} = 1 \Rightarrow y = \pm 2$ The graph is thus an ellipse centered at the origin, with intercepts at approximately (-4.24, 0), (4.24, 0), (0, 2), and (0, -2).