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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 coordinate... 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

2.1.1 Write down the coordinates of G.

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Answer

The coordinates of G can be determined by knowing that FG is parallel to the y-axis, meaning that G must share the same x-coordinate as F, which is -1. Since G lies on the circle defined by the equation $x^2 + y^2 = r^2$ , we can find y by substituting x = -1 into this equation:

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

Here, $r^2 = 26$ from the radius calculated earlier. Thus:

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

Step 2

2.1.2 Determine: (a) The gradient of OF

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Answer

The gradient of a line is calculated using the formula:

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

Here, O is the origin (0, 0) and F is (-1, 5). Substituting in:

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

Thus, the gradient of OF is -5.

Step 3

2.1.2 Determine: (b) The equation of the tangent to the circle at F in the form y = …

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Answer

To find the equation of the tangent line at point F(-1, 5), we use the point-slope form of a line, which is given by:

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

Using the gradient we calculated earlier (-5) and point F:

$y - 5 = -5(x - (-1))$

This simplifies to:

$y - 5 = -5(x + 1) \rightarrow y - 5 = -5x - 5 \rightarrow y = -5x + 0$

Thus, the equation of the tangent line at F is:

$y = -5x$ .

Step 4

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

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Answer

The equation $\frac{x^2}{7} + \frac{y^2}{64} = 1$ represents an ellipse centered at the origin.

Identify the intercepts:
- x-intercepts occur when $y = 0$ : $\frac{x^2}{7} + 0 = 1 \Rightarrow x^2 = 7 \Rightarrow x = \pm \sqrt{7} \approx \pm 2.65$
- y-intercepts occur when $x = 0$ : $0 + \frac{y^2}{64} = 1 \Rightarrow y^2 = 64 \Rightarrow y = \pm 8$
Drawing the graph:
- Plot the intercepts at approximately (-2.65, 0), (2.65, 0), (0, 8), and (0, -8).
- Sketch the ellipse connecting these points, ensuring the graph is symmetrical about both axes.

The resulting shape will be an elongated ellipse, indicating the relationship defined by the equation.

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

2.1.1 Write down the coordinates of G.

2.1.2 Determine: (a) The gradient of OF

2.1.2 Determine: (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: $$\frac{x^2}{7} + \frac{y^2}{64} = 1$$

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