In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin. FG is parallel to the y-axis. 2.1.1 Write down the coordinates of G. 2.1.2 Determine: (a) The gradient of OF (b) The equation of the tangent to the circle at F in the form y = ... 2.2 Draw, on the grid provided in the ANSWER BOOK, the graph defined by: $ \frac{x^2}{7} + \frac{y^2}{64} = 1 $ Clearly show ALL the intercepts with the axes.

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In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin - NSC Technical Mathematics - Question 2 - 2022 - Paper 2

Question 2

In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin. FG is parallel to the y-axis. 2.1.1 Write down the coordinates ... show full transcript

Worked Solution & Example Answer:In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin - NSC Technical Mathematics - Question 2 - 2022 - Paper 2

Step 1

Write down the coordinates of G.

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Answer

Given that G (x; y) lies on the circle centered at the origin, we can determine its coordinates based on F being at (-1, 5). Since FG is parallel to the y-axis, G will have the same y-coordinate as F, thus:

The coordinates of G are G (−1; y).
To find y, we can use the equation of the circle, derived from its radius:

$x^2 + y^2 = r^2$

Substituting F's coordinates gives us radius squared:

$r^2 = (-1)^2 + (5)^2 = 1 + 25 = 26.$

Now substituting x = -1 in the circle's equation:

$(-1)^2 + y^2 = 26$

ightarrow y^2 = 25 ightarrow y = 5 ext{ or } y = -5.$$ Thus, the coordinates of G are G (−1; 5) or G (−1; -5).

Step 2

Determine: The gradient of OF

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Answer

To find the gradient (m) of line segment OF (where O is the origin (0,0) and F is (-1, 5)), we use the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting O (0, 0) as (x_1, y_1) and F (−1, 5) as (x_2, y_2), we get:

$m = \frac{5 - 0}{-1 - 0} = \frac{5}{-1} = -5.$

Thus, the gradient of OF is -5.

Step 3

Determine: The equation of the tangent to the circle at F in the form y = ...

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Answer

Using the point-slope form of a line equation, which is:

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

Where (x_1, y_1) are the coordinates of point F and m is the gradient. Using the coordinates F (−1, 5) and the gradient as -\frac{1}{5}, we substitute:

$y - 5 = -\frac{1}{5}(x + 1).$

Simplifying:

ightarrow y = -\frac{1}{5}x + 5 - \frac{1}{5} ightarrow y = -\frac{1}{5}x + \frac{24}{5}.$$ Thus, the equation of the tangent at F is: $$y = -\frac{1}{5}x + \frac{24}{5}.$$

Step 4

Draw, on the grid, the graph defined by: $ \frac{x^2}{7} + \frac{y^2}{64} = 1 $

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Answer

Identify the major and minor axes:
- The equation indicates the ellipse has a semi-major axis of 8 (since ( \sqrt{64} = 8 )) and a semi-minor axis of approximately 2.645 (since ( \sqrt{7} \approx 2.645 )).
Plot the intercepts:
- For the x-intercepts, set y = 0:
$\frac{x^2}{7} = 1 \implies x^2 = 7 \implies x = \pm\sqrt{7}.$
- For the y-intercepts, set x = 0:
$\frac{y^2}{64} = 1 \implies y^2 = 64 \implies y = \pm8.$
Draw the ellipse, ensuring symmetry about both axes and including all intercepts marked clearly on the graph.

In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin - NSC Technical Mathematics - Question 2 - 2022 - Paper 2

Worked Solution & Example Answer:In the diagram below, F (−1; 5) and G (x; y) are points on the circle with the centre at the origin - NSC Technical Mathematics - Question 2 - 2022 - Paper 2

Write down the coordinates of G.

Determine: The gradient of OF

Determine: The equation of the tangent to the circle at F in the form y = ...

Draw, on the grid, the graph defined by: \( \frac{x^2}{7} + \frac{y^2}{64} = 1 \)

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