In the diagram below, A, B(-5; 12) and C(t; 0) lie on a circle with centre O at the origin - NSC Technical Mathematics - Question 2 - 2021 - Paper 2

Given that point C lies on the x-axis at (t, 0), we can substitute y = 0 into the circle's equation:

$x^2 + 0^2 = 169 \implies x^2 = 169 \implies x = \pm 13$

Since A is on the x-axis, t can take the value of either -13 or 13. As point C is on the positive side of the x-axis, we take:

$t = 13$

To find the equation of the tangent line at point B(-5, 12), we first need to determine the slope (m) of the radius OB, as the tangent at any point is perpendicular to the radius at that point. The coordinates for point O are (0, 0):

The slope of OB is:

$m_{OB} = \frac{12 - 0}{-5 - 0} = -\frac{12}{5}$

The slope of the tangent line (m_t) at point B will be the negative reciprocal:

$m_t = \frac{5}{12}$

Now using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substituting in point B for (x_1, y_1) and the slope:

$y - 12 = \frac{5}{12}(x + 5)$

Expanding this gives:

$y - 12 = \frac{5}{12}x + \frac{25}{12}$

Thus, rearranging to form y = ... yields:

$y = \frac{5}{12}x + \frac{169}{12}$

The given equation represents an ellipse centered at the origin. The semi-major axis (vertical) and semi-minor axis (horizontal) can be found by:

• Semi-major axis = (\sqrt{35}) along the y-axis
• Semi-minor axis = (\sqrt{16} = 4) along the x-axis

The intercepts on the axes:

• For the x-intercepts (when y=0): Set (y=0)

$\frac{x^2}{16} = 1 \implies x = \pm 4$

• For the y-intercepts (when x=0): Set (x=0)

$\frac{y^2}{35} = 1 \implies y = \pm \sqrt{35}$

Thus, the intercepts are (-4, 0), (4, 0), (0, \sqrt{35}), and (0, -\sqrt{35}). Illustrate these points on the grid provided.