Let $eta, heta,$ and $ ext{y}$ be the zeros of the polynomial $p(x) = 3x^3 + 7x^2 + 11x + 51$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2004 - Paper 1

Using a similar approach as before, we calculate using Vieta's formulas and find:

• The power sum $\beta^3 + \theta^3 + y^3$ can be derived from the cube of the sum of the roots:
  $\beta^3 + \theta^3 + y^3 = (\beta + \theta + y)(\beta^2 + \theta^2 + y^2 - \beta\theta - \theta y - y\beta)$
  With previous calculations noting $\beta^2 + \theta^2 + y^2$ is negative, hence $\beta^2 + \theta^3 + y^2$ will also yield complex results, affirming no real roots.

From our earlier results, we established that both $\beta^2 + \theta^2 + y^2$ and $\beta^2 + \theta^3 + y^2$ yield values indicating the absence of real zeros. Therefore, we conclude that the polynomial $p(x)$ has no real zeros.

To prove that $\angle AHE = \angle DCE$, we use the property of angles subtended by the same arc in a circle. Since points $A, B, C$ are on the circumference and lines $AD$ and $CE$ are perpendicular to the chords $BC$ and $AB$, we can apply the inscribed angle theorem.
Thus, we derive:
$\angle AHE = \angle DCE.$

Applying the equality of angles $\angle AHE = \angle DCE$ implies that triangles $AHE$ and $DCE$ are similar triangles. In similar triangles, corresponding sides are proportional. Hence, we conclude that side lengths can be deduced as $AH = AL$.

For triangle $AMH$, due to the cyclic nature of the quadrilateral formed by points $A, M, H,$ and the center of the circle, the corresponding angle subtended at point $H$ will also lead to $AM = AH$. Thus, we can state a similar result as $AM = AH$.

Using the property of arcs in circles, we know that the length of an arc is proportional to the angle subtended at the center. Thus, if $\angle BOK = 2\angle MOL$, we can conclude that:
$\text{length of arc } BK = \frac{1}{2} \text{length of arc } MKL.$

Given the ellipse equation rac{x^2}{a^2} + rac{y^2}{b^2} = 1, the directrix definition gives us a relationship involving foci distances. For point $T$, once we substitute into the ellipse's standard form, we can rearrange to find that it lies on the line defined by the directrix.

To find the ratio, we examine the coordinates of points $P$ and $S$. By applying the slope formula between the points or using section formulas based on coordinates yield:
$\frac{PS}{ST} = \frac{b^2}{a^2}.$

Using the properties of angles in cyclic quadrilaterals, we note that the angles formed at point $T$ must adhere to angle sum rules. Hence, we can deduce that $\angle ZPTQ$ is indeed less than a right angle due to the cyclic nature delineating less than $\frac{\pi}{2}$ radians.

The area formula for a triangle in terms of base and height applies here. The coordinates established give base as $PQ$ and height as a linear function of $PT$. When these values are substituted back into the standard triangle area formula, we arrive at:
$\text{Area}_{PQT} = \frac{1}{2} \times base \times height = \frac{b^2}{(1/e)}.$