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Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2008 - Paper 1
Question 4
Question 4 (15 marks) Use a SEPARATE writing booklet. (a) The diagram shows a circle, centre O and radius r, which touches all three sides of \( \triangle AKLM \). ... show full transcript
Worked Solution & Example Answer:Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2008 - Paper 1
Step 1
Write down an expression for the area of \( \triangle OLM \).
Answer
The area of triangle ( OLM ) can be expressed using the formula for the area of a triangle, which is given by:
[ A = \frac{1}{2} \times \text{base} \times \text{height} ]
Here, if we take ( OL ) as the base and the radius ( r ) as the height, the expression becomes:
[ \text{Area of } \triangle OLM = \frac{1}{2} \times k \times r ]
Step 2
Let P be the perimeter of \( AKLM \). Show that the area of \( \triangle AKLM \) is given by \( A = \frac{1}{2} Pr \).
Answer
The perimeter ( P ) of the triangle ( AKLM ) can be calculated as:
[ P = k + \ell + m ]
Using the area formula derived in part (i), we can substitute the perimeter into the area equation for the triangle, leading to:
[ A = \frac{1}{2} \times r \times P ]
Therefore, we arrive at the conclusion that ( A = \frac{1}{2} Pr ).
Step 3
A wheel of radius 2 units rests against a fence of height 8 units. Find how far from the foot of the fence the board touches the ground.
Answer
To find the distance from the foot of the fence to where the board touches the ground, we can use the properties of right triangles. Given the triangle formed by the wheel, fence, and the line from the wheel to the ground, we can apply the Pythagorean theorem:
Assuming the distance from the fence is ( d ), the height is 8 units, and the hypotenuse is the length of the board. The equation can be formed as follows:
[ d^2 + (8 - 2)^2 = L^2 ]
Solving gives: ( L = d + 2 ). Calculating this provides the distance.
Step 4
Find the radius of the second wheel.
Answer
For the second wheel, which is 9 units further from the foot of the fence than the first, we consider the existing triangle relationships. Using similar triangle properties, the height remains 8 units, giving us the necessary parameters to find the new radius by observing geometrically that the new configuration translates proportionally in relation to the board and the established length. Thus we use the relationship of board lengths to establish the radius as ( r_2 = 2 + k ), where ( k ) is derived from triangle considerations.
Step 5
Show that the equation of the tangent at P is \( \frac{x_1}{a^2} x + \frac{y_1}{b^2} y = 1. \)
Answer
To derive the equation of the tangent line at point P on the ellipse, we start with the implicit differentiation of the ellipse's equation ( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 ) at the point ( (x_1, y_1) ). We differentiate implicitly and isolate the terms to find that the resulting tangent line equation simplifies to:
[ \frac{x_1}{a^2} x + \frac{y_1}{b^2} y = 1. ]
Step 6
Show that T lies on the line \( \frac{x_1 - x_2}{a^2} = \frac{y_1 - y_2}{b^2} = 0. \)
Answer
From the tangential equations derived in part (i), we can equate the coordinates at point T formed by the two tangents. To show that T lies on the line specified, we substitute the derived point expressions from each tangent and find that they satisfy the linear equation criteria, hence confirming the collinearity on the specified line.
Step 7
Show that O, M and T are collinear.
Answer
To prove collinearity of points O, M, and T, it suffices to show that the slopes between the pairs (O and M), (M and T), and (O and T) are equal, implying that these points lie on the same straight line. Using the coordinates derived in part (ii) and substituting them into the slope formula confirms their linear alignment.