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The diagram shows two circles, C_1 and C_2, centred at the origin with radii a and b, where a > b - HSC - SSCE Mathematics Extension 2 - Question 5 - 2010 - Paper 1
Question 5
The diagram shows two circles, C_1 and C_2, centred at the origin with radii a and b, where a > b. The point A lies on C_1 and has coordinates (a \, \cos \theta, a ... show full transcript
Worked Solution & Example Answer:The diagram shows two circles, C_1 and C_2, centred at the origin with radii a and b, where a > b - HSC - SSCE Mathematics Extension 2 - Question 5 - 2010 - Paper 1
Step 1
(i) Write down the coordinates of B.
Answer
To find the coordinates of point B, we need to determine where the line OA intersects the circle C_2. The equation for circle C_2 is given by:
Since A is at (a \cos \theta, a \sin \theta), the line OA has the equation: y = \frac{\sin \theta}{\cos \theta} x = \tan \theta , x.
Substituting this line's equation into the equation of the circle gives:
From this we can find the x-coordinate of B, and substitute back to find y. Therefore, point B has coordinates: ( B \left( b \cos \theta, b \sin \theta \right) ).
Step 2
(ii) Show that P lies on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Answer
Point P is the intersection of the horizontal line through B and the vertical line through A. The coordinates of point P will be (a \cos \theta, b \sin \theta).
To show that P lies on the ellipse, we can substitute these coordinates into the ellipse equation: [ \frac{(a \cos \theta)^2}{a^2} + \frac{(b \sin \theta)^2}{b^2} = \frac{a^2 \cos^2 \theta}{a^2} + \frac{b^2 \sin^2 \theta}{b^2} = \cos^2 \theta + \sin^2 \theta = 1. ]
Hence, point P indeed lies on the ellipse.
Step 3
(iii) Find the equation of the tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at P.
Answer
The equation of the tangent to the ellipse at point (x_0, y_0) is given by: [ \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1. ]
For our point P(a \cos \theta, b \sin \theta), substituting these into the tangent equation yields: [ \frac{(a \cos \theta) x}{a^2} + \frac{(b \sin \theta) y}{b^2} = 1. ]
Thus, the equation of the tangent at P becomes: [ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1. ]
Step 4
(iv) Assume that A is not on the y-axis. Show that the tangent to the circle C_1 at A, and the tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at P, intersect at a point on the x-axis.
Answer
The tangent to circle C_1 at point A is described by the equation: [ x \cos \theta + y \sin \theta = a. ]
Substituting y = 0 (for the x-axis), we find: [ \cos \theta , x = a \implies x = \frac{a}{\cos \theta}. ]
Now, substituting y = 0 into the tangent of the ellipse at point P gives: [ \frac{a}{\cos \theta} \cos \theta/a + 0 = 1. ]
Both equations will hold true, confirming that their intersection lies on the x-axis.