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Consider the ellipse $E$ with equation \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\] and the points $P(a\cos\theta, b\sin\theta)$, $Q(a\cos(\theta + \phi), b\sin(\theta + \phi))$ and $R(a\cos(\theta - \phi), b\sin(\theta - \phi)$ on $E$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2001 - Paper 1
Question 5
![Consider-the-ellipse-$E$-with-equation--\[\frac{x^2}{a^2}-+-\frac{y^2}{b^2}-=-1.\]--and-the-points-$P(a\cos\theta,-b\sin\theta)$,-$Q(a\cos(\theta-+-\phi),-b\sin(\theta-+-\phi))$-and-$R(a\cos(\theta---\phi),-b\sin(\theta---\phi)$-on-$E$-HSC-SSCE Mathematics Extension 2-Question 5-2001-Paper 1.png](https://cdn.simplestudy.io/assets/backend/uploads/question/subject_1096_paper_15742_q5_67b8e802.jpg)
Consider the ellipse $E$ with equation \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\] and the points $P(a\cos\theta, b\sin\theta)$, $Q(a\cos(\theta + \phi), b\sin(\the... show full transcript
Worked Solution & Example Answer:Consider the ellipse $E$ with equation \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\] and the points $P(a\cos\theta, b\sin\theta)$, $Q(a\cos(\theta + \phi), b\sin(\theta + \phi))$ and $R(a\cos(\theta - \phi), b\sin(\theta - \phi)$ on $E$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2001 - Paper 1
Step 1
Show that the equation of the tangent to $E$ at the point $P$ is \[ \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1. \]
Answer
To find the equation of the tangent to the ellipse at point , we use the formula for the tangent line to an ellipse. The formula is given by:
[\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1.]
Plugging in the coordinates of point verifies that this indeed passes through .
Step 2
Show that the chord $QR$ is parallel to the tangent at $P$.
Answer
First, find the slope of the tangent at point . The slope can be derived from the tangent equation:
[y = -\frac{b\cos\theta}{a\sin\theta}x + b\sin\theta + \frac{b\cos\theta}{a}\cos\theta.]
Now, evaluate the coordinates of points and :
[Q = (a\cos(\theta + \phi), b\sin(\theta + \phi))] [R = (a\cos(\theta - \phi), b\sin(\theta - \phi))]
The slope of chord can be calculated as:
[\text{slope}_{QR} = \frac{y_Q - y_R}{x_Q - x_R}.]
If it is found to be equal to the slope of the tangent, we conclude that the chord is parallel to the tangent at .
Step 3
Deduce that $OP$ bisects the chord $QR$.
Answer
To prove that bisects the chord , we utilize the fact that the midpoint of will lie on line . If the slopes of the two lines are equal, and knowing that they intersect at point , we can show by symmetry that divides into two equal parts. Therefore, it can be deduced that bisects the chord.