Question 10
The curve $C$ with equation $y=\frac{6-3x}{x-4}$ where $x \neq 4$, crosses the $x$-axis at the point $P$ and the $y$-axis at the point $Q$(a) Find the coordinates of
(i) $P$
(ii) $Q$ [ 2 Marks ]
(b) Write down an equation of the asymptote to $C$ which is
(i) parallel to the $y$-axis
(ii) parallel to the $x$-axis [2 Marks ]
(c) Sketch $C$ showing clearly the asymptotes and the coordinates of the points $P$ and $Q$ [ 3 Marks ]
The line $L$ is the normal to $C$ at the point on $C$ where $x=2$
(d) Find an equation of $L$ [ 6 Marks ]
The line $L$ intersects $C$ again at the point $R$
(e) Find the $x$ coordinate of $R$ [ 3 Marks ]
Solution
The given curve $C$ has the equation: $$ y=\frac{6-3x}{x-4}, \quad x \neq 4. $$(a) Coordinates of Points
(i) Point $P$ (Where $C$ crosses the $x$-axis)
At the $x$-axis, $y=0$. Substituting $y=0$ into the equation: $$ 0=\frac{6-3x}{x-4}. $$ For the fraction to be zero, the numerator must be zero: $$ 6-3x=0 \implies x=2. $$ Thus, the coordinates of $P$ are: $$ P=(2,0). $$(ii) Point $Q$ (Where $C$ crosses the $y$-axis)
At the $y$-axis, $x=0$. Substituting $x=0$ into the equation: $$ y=\frac{6-3(0)}{0-4}=\frac{6}{-4}=-\frac{3}{2}. $$ Thus, the coordinates of $Q$ are: $$ Q=\left(0,-\frac{3}{2}\right). $$(b) Asymptotes of $C$
(i) Asymptote Parallel to the $y$-axis
The denominator of the equation $$ y=\frac{6-3x}{x-4} $$ becomes zero when $$ x-4=0, $$ i.e., $$ x=4. $$ Thus, the vertical asymptote is: $$ x=4. $$(ii) Asymptote Parallel to the $x$-axis
As $x \to \infty$ or $x \to -\infty$, the dominant term in the numerator and denominator is $-3x$ and $x$, respectively. Simplifying: $$ y \to -3. $$ Thus, the horizontal asymptote is: $$ y=-3. $$(c) Sketch of the Curve $C$
The curve passes through $P(2,0)$ and $Q\left(0,-\frac{3}{2}\right)$, has a vertical asymptote at $x=4$, and a horizontal asymptote at $y=-3$. The sketch should clearly show these features.
(d) Equation of the Normal Line $L$ at $x=2$
First, find the slope of the tangent to $C$ at $x=2$. The derivative of $$ y=\frac{6-3x}{x-4} $$ is computed using the quotient rule: $$ y'=\frac{(x-4)(-3)-(6-3x)(1)}{(x-4)^2}. $$ Simplify: $$ y'=\frac{-3(x-4)-(6-3x)}{(x-4)^2} =\frac{-3x+12-6+3x}{(x-4)^2} =\frac{6}{(x-4)^2}. $$ At $x=2$, the slope of the tangent is: $$ y'(2)=\frac{6}{(2-4)^2} =\frac{6}{4} =\frac{3}{2}. $$ The slope of the normal is the negative reciprocal: $$ \text{slope of normal}=-\frac{2}{3}. $$ The point on $C$ at $x=2$ is $P(2,0)$. Using the point-slope form of a line: $$ y-0=-\frac{2}{3}(x-2). $$ Simplify: $$ y=-\frac{2}{3}x+\frac{4}{3}. $$ Thus, the equation of the normal line is: $$ L:y=-\frac{2}{3}x+\frac{4}{3}. $$(e) Intersection of $L$ and $C$
Substitute $$ y=-\frac{2}{3}x+\frac{4}{3} $$ into $$ y=\frac{6-3x}{x-4}: $$ $$ -\frac{2}{3}x+\frac{4}{3} = \frac{6-3x}{x-4}. $$ Clear the fractions by multiplying through by $3(x-4)$: $$ -2x(x-4)+4(x-4)=3(6-3x). $$ Expand: $$ -2x^2+8x+4x-16=18-9x. $$ Simplify: $$ -2x^2+12x-16=18-9x. $$ Rearrange to form a quadratic equation: $$ -2x^2+21x-34=0. $$ Divide through by $-1$: $$ 2x^2-21x+34=0. $$ Solve this quadratic using the quadratic formula: $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, \quad a=2,\ b=-21,\ c=34. $$ $$ x=\frac{-(-21)\pm\sqrt{(-21)^2-4(2)(34)}}{2(2)}. $$ $$ x=\frac{21\pm\sqrt{441-272}}{4}. $$ $$ x=\frac{21\pm\sqrt{169}}{4}. $$ $$ x=\frac{21\pm13}{4}. $$ $$ x=\frac{34}{4}=8.5 \quad \text{or} \quad x=\frac{8}{4}=2. $$ Since $x=2$ corresponds to the point $P$, the other intersection point is: $$ x=8.5. $$Question 11
The roots of a quadratic equation $E$ are $\alpha$ and $\beta$ where $\alpha>\beta>0$. Given that $\alpha-\beta=2\sqrt{6}$ and $\alpha^2+\beta^2=30$(a) show that
(i) $\alpha\beta=3$ [ 3 Marks ]
(ii) $\alpha+\beta=6$ [ 2 Marks ]
(b) Without solving $E$
(i) find the value of $\alpha^4+\beta^4$ [ 2 Marks ]
(ii) find the exact value of $\alpha^4-\beta^4$ [ 2 Marks ]
Given that $\alpha^4=P+Q\sqrt{6}$ where $P$ and $Q$ are positive integers,
(c) find the value of $P$ and the value of $Q$ [ 2 Marks ]
Solution
The roots of the quadratic equation $E$ are $\alpha$ and $\beta$, where $\alpha>\beta>0$.We are given: $$ \alpha-\beta=2\sqrt{6}, \quad \alpha^2+\beta^2=30. $$
(a) Show that
(i) $\alpha\beta=3$
Method 1
We know that:
$$
\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.
$$
Substitute the given values:
$$
30=(\alpha+\beta)^2-2\alpha\beta.
$$
Let $s=\alpha+\beta$ and $p=\alpha\beta$. Then:
$$
30=s^2-2p. \quad \dots (1)
$$
We also know:
$$
\alpha-\beta=2\sqrt{6}.
$$
$$
\alpha+\beta=s, \quad \alpha-\beta=2\sqrt{6}.
$$
Using the identity:
$$
\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta,
$$
and substituting $(\alpha-\beta)^2=24$, we can solve for $\alpha\beta$:
$$
30=s^2-2p, \quad \text{and by substitution } s=6:
$$
$$
30=6^2-2p.
$$
$$
30=36-2p.
$$
$$
2p=6 \implies p=3.
$$
Thus, we have:
$$
\alpha\beta=3.
$$
Method 2
Given:
$$
\alpha-\beta=2\sqrt{6}, \quad \alpha^2+\beta^2=30.
$$
$$
(\alpha-\beta)^2=(2\sqrt{6})^2
\Rightarrow
\alpha^2+\beta^2-2\alpha\beta=24
\Rightarrow
24=30-2\alpha\beta
$$
$$
\Rightarrow
2\alpha\beta=30-24
\Rightarrow
\alpha\beta=3
$$
(ii) $\alpha+\beta=6$
From the above, we already deduced that $$ s=\alpha+\beta=6. $$ Thus: $$ \alpha+\beta=6. $$(b) Without solving $E$
(i) Find $\alpha^4+\beta^4$
Using the identity: $$ \alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2, $$ substitute the known values: $$ \alpha^2+\beta^2=30, \quad \alpha\beta=3. $$ $$ \alpha^4+\beta^4=30^2-2(3^2). $$ $$ \alpha^4+\beta^4=900-18=882. $$ Thus: $$ \alpha^4+\beta^4=882. $$(ii) Find $\alpha^4-\beta^4$
Using the identity: $$ \alpha^4-\beta^4=(\alpha^2+\beta^2)(\alpha^2-\beta^2), $$ we already know $$ \alpha^2+\beta^2=30. $$ To find $\alpha^2-\beta^2$, use the identity: $$ \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta). $$ Substitute $\alpha+\beta=6$ and $\alpha-\beta=2\sqrt{6}$: $$ \alpha^2-\beta^2=6(2\sqrt{6})=12\sqrt{6}. $$ Now substitute: $$ \alpha^4-\beta^4=30\cdot12\sqrt{6}. $$ $$ \alpha^4-\beta^4=360\sqrt{6}. $$(c) Express $\alpha^4=P+Q\sqrt{6}$
We are given $$ \alpha^4=P+Q\sqrt{6}, $$ where $P$ and $Q$ are positive integers. From part (b): $$ \alpha^4+\beta^4=882, \quad \alpha^4-\beta^4=360\sqrt{6}. $$ Using the identities: $$ \alpha^4= \frac{(\alpha^4+\beta^4)+(\alpha^4-\beta^4)}{2}, $$ $$ \beta^4= \frac{(\alpha^4+\beta^4)-(\alpha^4-\beta^4)}{2}. $$ Substitute the values: $$ \alpha^4= \frac{882+360\sqrt{6}}{2}, \quad \beta^4= \frac{882-360\sqrt{6}}{2}. $$ $$ \alpha^4=441+180\sqrt{6}. $$ Thus: $$ P=441, \quad Q=180. $$Final Answers
- (a) (i) $\alpha\beta=3$, (ii) $\alpha+\beta=6$.
- (b) (i) $\alpha^4+\beta^4=882$, (ii) $\alpha^4-\beta^4=360\sqrt{6}$.
- (c) $P=441$, $Q=180$.
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