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 $E$. (i) Show that the equation of the tangent to $E$ at the point $P$ is \[\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1.\] (ii) Show that the chord $QR$ is parallel to the tangent at $P$. (iii) Deduce that $OP$ bisects the chord $QR$.

SSCE HSC Mathematics Extension 2 Resisted horizontal motion: Consider the ellipse $E$ with eq

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$

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 \[ \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1. \]

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Answer

To find the equation of the tangent to the ellipse at point $P(a\cos\theta, b\sin\theta)$ , 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 $P$ verifies that this indeed passes through $P$ .

Step 2

Show that the chord $QR$ is parallel to the tangent at $P$.

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Answer

First, find the slope of the tangent at point $P$ . 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 $Q$ and $R$ :

[Q = (a\cos(\theta + \phi), b\sin(\theta + \phi))] [R = (a\cos(\theta - \phi), b\sin(\theta - \phi))]

The slope of chord $QR$ 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 $QR$ is parallel to the tangent at $P$ .

Step 3

Deduce that $OP$ bisects the chord $QR$.

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Answer

To prove that $OP$ bisects the chord $QR$ , we utilize the fact that the midpoint of $QR$ will lie on line $OP$ . If the slopes of the two lines are equal, and knowing that they intersect at point $P$ , we can show by symmetry that $OP$ divides $QR$ into two equal parts. Therefore, it can be deduced that $OP$ bisects the chord.

Show that the equation of the tangent to $E$ at the point $P$ is \[ \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1. \]

Show that the chord $QR$ is parallel to the tangent at $P$.

Deduce that $OP$ bisects the chord $QR$.

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