Question: Line $y = x - 1$ meets curve $y^2 = 9x$ at points $A$ and $B$. Find the exact length $AB$ in surd form.

Step 1. Substitute the line into the curve: $(x-1)^2 = 9x$:

$$x^2 - 2x + 1 = 9x \Rightarrow x^2 - 11x + 1 = 0$$

Step 2. Solve for $x$: $x = \dfrac{11 \pm \sqrt{121 - 4}}{2} = \dfrac{11 \pm \sqrt{117}}{2}$.

Let $x_A = \dfrac{11 - \sqrt{117}}{2}$, $x_B = \dfrac{11 + \sqrt{117}}{2}$. Corresponding $y_A = x_A - 1$, $y_B = x_B - 1$.

Step 3. Distance $AB$:

$$AB^2 = (x_B - x_A)^2 + (y_B - y_A)^2 = 2(x_B - x_A)^2$$

(since $y_B - y_A = x_B - x_A$).

$x_B - x_A = \sqrt{117}$.

$$AB^2 = 2 \cdot 117 = 234 \Rightarrow AB = \sqrt{234} = 3\sqrt{26}$$

Answer: $AB = 3\sqrt{26}$.

Tip: When you have line $y = mx + c$ meeting a curve, the chord length is:

$$AB = |x_B - x_A| \cdot \sqrt{1 + m^2}$$

This bypasses computing $y$-values. For $m = 1$: factor $\sqrt{2}$.

Question: Find the set of values of $k$ for which $y + 1 = kx$ meets $3 - xy + x^2 = 0$ at two distinct points.

Step 1. From line: $y = kx - 1$. Substitute into curve:

$$3 - x(kx - 1) + x^2 = 0$$ $$3 - kx^2 + x + x^2 = 0$$ $$(1 - k)x^2 + x + 3 = 0$$

Step 2. For two distinct intersections, discriminant $> 0$ AND coefficient of $x^2$ nonzero.

Coefficient $1 - k \neq 0 \Rightarrow k \neq 1$.

Discriminant: $1 - 12(1 - k) > 0 \Rightarrow 1 - 12 + 12k > 0 \Rightarrow 12k > 11 \Rightarrow k > \dfrac{11}{12}$.

Step 3. Combine: $k > \dfrac{11}{12}$ and $k \neq 1$.

Answer: $k \in \left(\dfrac{11}{12}, 1\right) \cup (1, \infty)$.

Tip: When the substituted equation has a parameter in the LEADING coefficient, check that case separately ($k = 1$ degenerates the quadratic to a linear equation).

Question: The line $y = mx + 1$ and curve $y = 2x^2 - x + n$ intersect at the point $(1, 4)$. Find $m$ and $n$.

Step 1. Both equations must hold at $(1, 4)$:

From line: $4 = m(1) + 1 \Rightarrow m = 3$.

From curve: $4 = 2(1)^2 - 1 + n = 1 + n \Rightarrow n = 3$.

Step 2. Verify: line is $y = 3x + 1$, curve is $y = 2x^2 - x + 3$. Setting equal:

$$3x + 1 = 2x^2 - x + 3 \Rightarrow 2x^2 - 4x + 2 = 0 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1$$

So the line is actually TANGENT to the curve at $(1, 4)$. $\checkmark$

Answer: $m = 3$, $n = 3$.

Tip: "Curve passes through point" or "line meets curve at point": substitute the point's coordinates into the equation(s). Each unknown needs one equation.

Question: Solve $27^y \cdot 3^x = 1$ and $2^{y-2} \div 4^x = 32^{1/y}$.

Step 1. Take $\log$ (or rewrite all terms with common base).

For the first equation: $27 = 3^3$, so $27^y = 3^{3y}$:

$$3^{3y + x} = 3^0 \Rightarrow 3y + x = 0 \Rightarrow x = -3y \quad (*)$$

Step 2. For the second equation: $4 = 2^2$, $32 = 2^5$:

$$2^{y - 2 - 2x} = 2^{5/y}$$ $$y - 2 - 2x = \dfrac{5}{y} \quad (**)$$

Step 3. Sub $x = -3y$ into $(**)$:

$$y - 2 - 2(-3y) = \dfrac{5}{y}$$ $$y - 2 + 6y = \dfrac{5}{y}$$ $$7y - 2 = \dfrac{5}{y}$$ $$7y^2 - 2y - 5 = 0$$ $$(7y + 5)(y - 1) = 0$$ $$y = 1 \text{ or } y = -\dfrac{5}{7}$$

Step 4. Find $x$:

• $y = 1$: $x = -3$.
• $y = -\tfrac{5}{7}$: $x = \dfrac{15}{7}$.

Answer: $(x, y) = (-3, 1)$ or $\left(\dfrac{15}{7}, -\dfrac{5}{7}\right)$.

Tip: Exponential simultaneous systems: reduce both sides to a common base, equate exponents. This converts the exponential system into a polynomial/rational system.

Question: Two resistors $R_1$ and $R_2$ (in parallel) satisfy $R_1 + R_2 = 500$ and $\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{120}$. Find $R_1$ and $R_2$.

Step 1. From the second equation, combine fractions:

$$\dfrac{R_2 + R_1}{R_1 R_2} = \dfrac{1}{120} \Rightarrow R_1 R_2 = 120(R_1 + R_2) = 120 \times 500 = 60000$$

Step 2. Now $R_1$ and $R_2$ are roots of the quadratic with:

• Sum $= 500$
• Product $= 60000$

By Vieta: $t^2 - 500t + 60000 = 0$.

Step 3. Solve: discriminant $= 250000 - 240000 = 10000 = 100^2$.

$$t = \dfrac{500 \pm 100}{2} = 300 \text{ or } 200$$

Answer: $\{R_1, R_2\} = \{300, 200\}$ (in $\Omega$).

Tip: "Sum and product" type word problems: use Vieta's formulas to construct a quadratic whose roots are the unknowns. $t^2 - (\text{sum})t + (\text{product}) = 0$.

Question: Line $y = -x + 4$ meets parabola $y^2 = 9x$. Find the coordinates of the intersection points.

Step 1. Substitute $y = -x + 4$ into $y^2 = 9x$:

$$(-x + 4)^2 = 9x$$ $$x^2 - 8x + 16 = 9x$$ $$x^2 - 17x + 16 = 0$$ $$(x - 1)(x - 16) = 0$$ $$x = 1 \text{ or } x = 16$$

Step 2. Find $y$-values:

• $x = 1$: $y = -1 + 4 = 3$.
• $x = 16$: $y = -16 + 4 = -12$.

Step 3. Verify with curve:

• $(1, 3)$: $3^2 = 9 = 9 \times 1$. $\checkmark$
• $(16, -12)$: $(-12)^2 = 144 = 9 \times 16$. $\checkmark$

Answer: $A = (1, 3)$, $B = (16, -12)$.

Tip: Line + conic: substitute the line's expression for $y$ (or $x$) into the conic, reduce to a single-variable polynomial, solve, then back-substitute.

Question: Line $x - y = 3$ meets curve $x^2 - 3xy + y^2 + 19 = 0$. Find the points of intersection.

Step 1. From line: $x = y + 3$. Substitute into the curve:

$$(y+3)^2 - 3(y+3)y + y^2 + 19 = 0$$ $$y^2 + 6y + 9 - 3y^2 - 9y + y^2 + 19 = 0$$ $$-y^2 - 3y + 28 = 0$$ $$y^2 + 3y - 28 = 0$$ $$(y + 7)(y - 4) = 0$$ $$y = -7 \text{ or } y = 4$$

Step 2. Find $x$-values:

• $y = -7$: $x = -4$.
• $y = 4$: $x = 7$.

Answer: $(-4, -7)$ and $(7, 4)$.

Tip: Curves with $xy$ cross-terms: substitute, expand carefully — cross-terms become polynomial pieces. Watch out for sign errors when distributing the $-3xy$ term.

Question: Line $x + y = k$ meets curve $y = \dfrac{9}{x} - 2$. For what values of $k$ do they have two distinct intersection points?

Step 1. Substitute: $x + \dfrac{9}{x} - 2 = k$. Multiply through by $x$ (noting $x \neq 0$):

$$x^2 + 9 - 2x = kx$$ $$x^2 - (k+2)x + 9 = 0$$

Step 2. Two distinct intersections: discriminant $> 0$:

$$(k+2)^2 - 36 > 0$$ $$(k+2)^2 > 36$$ $$k + 2 > 6 \text{ or } k + 2 < -6$$ $$k > 4 \text{ or } k < -8$$

Answer: $k > 4$ or $k < -8$.

Tip: For rational/reciprocal curves $y = \dfrac{a}{x} + b$, multiply through to clear the denominator BEFORE applying discriminant. Always check that the cleared-denominator value ($x = 0$ here) isn't a spurious solution.