Consider the ellipse $ ext{E}$ with equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, and the points $ P(acostheta, bsintheta) $, $ Q(acostheta + phi, bsin(theta + phi)) $, and $ R(acostheta - phi, bsin(theta - phi)) $ on $ \text{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 $.

Photo AI

Consider the ellipse $ ext{E}$ with equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, and the points $ P(acostheta, bsintheta) $, $ Q(acostheta + phi, bsin(theta + phi)) $, and $ R(acostheta - phi, bsin(theta - phi)) $ on $ \text{E} $ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2001 - Paper 1

Question 5

$Consider-the-ellipse-$-ext{E}$-with-equation-$-\frac{x^2}{a^2}-+-\frac{y^2}{b^2}-=-1-$,-and-the-points-$-P(acostheta,-bsintheta)-$,-$-Q(acostheta-+-phi,-bsin(theta-+-phi))-$,-and-$-R(acostheta---phi,-bsin(theta---phi))-$-on-$-\text{E}-$-HSC-SSCE Mathematics Extension 2-Question 5-2001-Paper 1.png$

Consider the ellipse $ ext{E}$ with equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, and the points $ P(acostheta, bsintheta) $, \( Q(acostheta + phi, bsin(th... show full transcript

Worked Solution & Example Answer:Consider the ellipse $ ext{E}$ with equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, and the points $ P(acostheta, bsintheta) $, $ Q(acostheta + phi, bsin(theta + phi)) $, and $ R(acostheta - phi, bsin(theta - phi)) $ on $ \text{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:

96%

114 rated

Answer

To find the equation of the tangent at the point ( P(acostheta, bsintheta) ) on the ellipse ( E ), we differentiate the equation of the ellipse implicitly.

Starting with the equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ we differentiate both sides with respect to ( x ) using implicit differentiation: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0.$ This gives us: $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}.$ Substituting ( P ) into this derivative gives the slope ( m ) at point P: $m = -\frac{b^2 \cdot a \cos \theta}{a^2 \cdot b \sin \theta} = -\frac{b \cos \theta}{a \sin \theta}.$ The equation of the tangent line can be found using point-slope form: $y - bsintheta = m(x - acostheta).$ Expanding and rearranging leads to: $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1.$

Step 2

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

99%

104 rated

Answer

To show that ( QR ) is parallel to the tangent at ( P ), we need to find the slopes of both segments. The slope of chord ( QR ) can be calculated from points ( Q ) and ( R ):

Coordinates of ( Q ) and ( R ):

( Q(acostheta + phi, bsin(theta + phi)) )
( R(acostheta - phi, bsin(theta - phi)) )

The slope ( m_{QR} ) is given by:

$m_{QR} = \frac{bsin(theta + phi) - bsin(theta - phi)}{(acostheta + phi) - (acostheta - phi)} = \frac{b[\sin(theta + phi) - \sin(theta - phi)]}{2phi}.$

Using the sine subtraction formula: $\sin(a + b) - \sin(a - b) = 2\cos(a)\sin(b),$ we get: $m_{QR} = \frac{b imes 2 \cos \theta \times \sin \phi}{2\phi} = \frac{b \cos \theta \sin \phi}{\phi}.$

Since the tangent at ( P ) has a slope of: $m = -\frac{b \cos \theta}{a \sin \theta}$ , we find that if both slope equations yield the same relationship in terms of trigonometric ratios, then chord ( QR ) is indeed parallel to the tangent at ( P ).

Step 3

Deduce that OP bisects the chord QR.

96%

101 rated

Answer

Since ( OP ) is perpendicular to the tangent at ( P ), and both ( QR ) and the tangent at ( P ) are parallel, it must follow by symmetry that the midpoint of chord ( QR ) lies directly above (or below) ( OP ). Thus, we conclude that ( OP ) bisects chord ( QR ). This follows from the properties of ellipses where the lines drawn from the center to the endpoints of a chord bisect that chord.

Show that the equation of the tangent to E at the point P is:

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

Deduce that OP bisects the chord QR.

Join the SSCE students using SimpleStudy...