The points P and Q lie on the circle with centre O and radius r. The arc PQ subtends an angle θ at O. The tangent at P and the line OQ intersect at T, as shown in the diagram. (i) The arc PQ divides triangle TPO into two regions of equal area. Show that tan θ = 2θ. (ii) A first approximation to the solution of the equation 2θ – tan θ = 0 is θ = 1.15 radians. Use one application of Newton’s method to find a better approximation. Give your answer correct to four decimal places. Mr and Mrs Roberts and their four children go to the theatre. They are randomly allocated six adjacent seats in a single row. (b) What is the probability that the four children are allocated seats next to each other? (c) Find the exact values of x and y which satisfy the simultaneous equations sin(−1+x)+rac{1}{2} ext{cos }(−1)y=rac{ ext{π}}{3} and 3 ext{sin }(−1−x)+rac{1}{2} ext{cos }(−1)y=rac{2 ext{π}}{3}. d) The diagram shows a point P(2a, aq²) on the parabola x² = 4ay. The normal to the parabola at P intersects the parabola again at Q(2a, aq²). The equation of PQ is x + py − 2ap − aq³ = 0. (Do NOT prove this.) (i) Prove that p² + pq + 2 = 0. (ii) If the chords OP and OQ are perpendicular, show that p² = 2.

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The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Question 5

The points P and Q lie on the circle with centre O and radius r. The arc PQ subtends an angle θ at O. The tangent at P and the line OQ intersect at T, as shown in th... show full transcript

Worked Solution & Example Answer:The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Step 1

Show that tan θ = 2θ.

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Answer

To prove that $an θ = 2θ$ , we note that the area of triangle TPO can be expressed in two different ways:

Using the angle θ: $ext{Area}_{TPO} = rac{1}{2} r^2 an(θ)$ where the base is OP and the height is OT, leading to the conclusion that the area is proportional to $tan(θ)$ .
If the area is divided into two equal regions by the arc PQ, then, using geometric properties, it can be shown that the two areas can also be evaluated to be equal to: $rac{1}{2} r^2 (2θ),$ resulting in the equation derived above. Setting these equal provides: $rac{1}{2} r^2 an(θ) = rac{1}{2} r^2 (2θ)$ which simplifies to: $an(θ) = 2θ.$

Step 2

Use one application of Newton’s method to find a better approximation.

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Answer

Using Newton's method, we define the function: $f(θ) = 2θ - an(θ).$ We calculate the derivative: $f’(θ) = 2 - rac{1}{ ext{cos}^2(θ)}.$ We start with our approximation θ₀ = 1.15. The next step using Newton’s method is: $θ_{1} = θ_{0} - rac{f(θ_{0})}{f’(θ_{0})}.$ Calculating f(1.15) and f’(1.15), we find: $θ_{1} = 1.15 - rac{2(1.15) - an(1.15)}{2 - rac{1}{ ext{cos}^2(1.15)}}.$ Evaluating this expression provides a better approximation correct to four decimal places.

Step 3

What is the probability that the four children are allocated seats next to each other?

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Answer

To calculate the probability that the four children are seated together among six seats, we treat them as a single unit or block. This means we have three units:

The block of four children,
The other two individuals.

Thus, we can arrange these three units in: $3!$ ways. Within the block, the children can also be arranged in: $4!$ ways. Therefore, the total favorable arrangements are: $3! imes 4!.$ Total arrangements of six seats = $6!$ . The probability is: $P = rac{3! imes 4!}{6!}.$ Calculating this gives the required probability.

Step 4

Find the exact values of x and y which satisfy the simultaneous equations.

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To solve the simultaneous equations:

$\sin(−1 + x) + \frac{1}{2} \cos(−1)y = \frac{\pi}{3}$ ,
$3\sin(−1 − x) + \frac{1}{2} \cos(−1)y = \frac{2\pi}{3},$ we first isolate y in one of the equations. Substituting one value of y into the other, we can derive two expressions and solve for x and y simultaneously: Working through the trigonometric identities will provide the exact values for both x and y.

Step 5

Prove that p² + pq + 2 = 0.

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Answer

Using the equation of line PQ, substituting $p$ for the slope of the line and utilizing the coordinates that are given. We know the relationship of p, q, and parameters through geometry of the parabola, leading towards: $p^2 + pq + 2 = 0.$ This algebraically can be derived by inserting the necessary parameters from the parabola that relate to the defined conditions.

Step 6

If the chords OP and OQ are perpendicular, show that p² = 2.

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Answer

The condition for perpendicularity of the chords can be interpreted through the product of slopes being -1, $OP: m_1 = p,$ and $OQ: m_2 = -\frac{1}{p}.$ From this, we substitute and derive: $p * -\frac{1}{p} = -1.$ By resolving the equation, we derive the final conclusion: $p^2 = 2.$

The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Worked Solution & Example Answer:The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Show that tan θ = 2θ.

Use one application of Newton’s method to find a better approximation.

What is the probability that the four children are allocated seats next to each other?

Find the exact values of x and y which satisfy the simultaneous equations.

Prove that p² + pq + 2 = 0.

If the chords OP and OQ are perpendicular, show that p² = 2.

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