FAD1014: MATHEMATICS II — Hard-Difficulty Simulation Paper

Academic Session 2025/2026 — Leak-Intensive Focus

[!warning] Difficulty Notice This paper targets difficulty level 7–8/10. Questions are designed to interleave topics, hide traps, and require multi-step reasoning. Part A builds foundation; Part B reaches full difficulty.
Duration 2 Hours
Total Marks 80
Instructions Answer ALL questions in Part A. Answer any FOUR of the six questions in Part B. Clear reasoning must be shown.

Formula Sheet (Provided)

Standard Integrals

$$\int x^n , dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$ $$\int e^{ax} , dx = \frac{e^{ax}}{a} + C$$ $$\int \frac{1}{x} , dx = \ln|x| + C$$ $$\int \sin x , dx = -\cos x + C, \quad \int \cos x , dx = \sin x + C$$ $$\int \sec^2 x , dx = \tan x + C, \quad \int \sec x \tan x , dx = \sec x + C$$

Summation & Standard Series

$$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$

Maclaurin & Binomial Series

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Function Series Valid for
$e^x$ $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$ All $x$
$\sin x$ $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ All $x$
$\cos x$ $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$ All $x$
$\ln(1+x)$ $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$ $-1 < x \leq 1$
$(1+x)^n$ $\displaystyle \sum_{r=0}^\infty \binom{n}{r} x^r$ $ x < 1$

Conic Sections

Shape Standard Equation Key Properties
Parabola $(y-k)^2 = 4a(x-h)$ Vertex: $(h,k)$, Focus: $(h+a,k)$, Directrix: $x = h-a$
Parabola $(x-h)^2 = 4a(y-k)$ Vertex: $(h,k)$, Focus: $(h,k+a)$, Directrix: $y = k-a$
Ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2 - b^2$ (horizontal), $c^2 = b^2 - a^2$ (vertical)
Hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2 + b^2$, Asymptotes: $y-k = \pm\frac{b}{a}(x-h)$

Identifying Differential Equation Types

Type Form Key Indicator
Separable $\displaystyle \frac{dy}{dx} = f(x)g(y)$ Variables can be split across $=$
Homogeneous $\displaystyle \frac{dy}{dx} = f!\left(\frac{y}{x}\right)$ Every term has same total degree
Bernoulli $\displaystyle \frac{dy}{dx} + P(x)y = Q(x)y^n$ $y^n$ factor on RHS, $n \neq 0, 1$
Exact $M(x,y),dx + N(x,y),dy = 0$ $\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Non-homogeneous $(a_1x + b_1y + c_1)dx + (a_2x + b_2y + c_2)dy = 0$ Constant terms present

PART A — Answer ALL Questions (24 Marks)

Question 1 [12 marks]

(a) Using the Maclaurin series definition (computing successive derivatives at $x = 0$), find the series expansion for $f(x) = \cos x$ up to and including the term in $x^4$.  [3]

(b) A curve is defined by the equation

$$9x^2 + 4y^2 - 18x + 16y - 11 = 0.$$

 (i) Show that the equation represents an ellipse by completing the square. Express it in standard form.  [3]

 (ii) Hence, determine the centre, vertices, and orientation of the ellipse.  [2]

(c) Solve the separable differential equation

$$\frac{dy}{dx} = \frac{2x}{y + 1}, \qquad y(0) = 1.$$

Express $y$ explicitly as a function of $x$.  [4]

Question 2 [12 marks]

(a) Write down the Maclaurin series for $\ln(1+u)$ up to the term in $u^4$, stating the range of $u$ for which it converges.

Hence, find the Maclaurin series for $f(x) = \ln(1 + 3x^2)$ up to the term in $x^6$, and state the range of $x$ for which this series is valid.  [3]

(b) Determine whether the sequence

$$a_n = \frac{5n^2 - 2n + 1}{2n^2 + 3n - 4}$$

converges or diverges. If it converges, find its limit.  [2]

(c) Using the method of differences, evaluate

$$\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}.$$

Hence, find the sum to infinity of this series.  [4]

(d) An ellipse has centre at $(2, -1)$, a vertex at $(2, 4)$, and a focus at $(2, 2)$. Determine the equation of this ellipse in standard form.  [3]

PART B — Answer Any FOUR Questions (56 Marks)

Question 3 [14 marks] — Differential Equations & Maclaurin Series (Interleaved)

(a) Show that the differential equation

$$\frac{dy}{dx} + y = xy^2$$

is a Bernoulli equation. State the values of $P(x)$, $Q(x)$, and $n$.  [2]

(b) Use the substitution $v = y^{-1}$ to reduce the equation to a first-order linear differential equation in $v$ and $x$.  [4]

(c) Solve the linear equation in $v$ using an integrating factor. Hence, find the general solution for $y$ in terms of $x$.  [5]

(d) Given the initial condition $y(0) = 1$, find the particular solution. Then, without expanding this solution directly, use the original differential equation to compute $y'(0)$, $y''(0)$, and $y'''(0)$. Hence, write the Maclaurin series for $y(x)$ up to the term in $x^3$.  [3]

Question 4 [14 marks] — Differential Equations & Ellipse (Interleaved)

A curve has the property that the gradient of its tangent at any point $P(x,y)$ satisfies

$$\frac{dy}{dx} = -\frac{9x}{4y}.$$

(a) Show that this is a separable differential equation and find its general solution.  [3]

(b) Given that the curve passes through the point $(2, 0)$, find the particular solution and rewrite it in the standard conic form $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.  [3]

(c) Identify the type of conic section represented by the solution. Determine its centre, vertices, and foci.  [4]

(d) Find the equation of the tangent to this curve at the point $\displaystyle \left(\frac{4}{3}, \sqrt{5}\right)$. Verify that this point lies on the curve.  [4]

Question 5 [14 marks] — Maclaurin Series & Ellipse (Interleaved)

(a) Write down the binomial expansion of $(1+u)^{-1/2}$ up to the term in $u^3$. State the range of $u$ for which the expansion is valid.  [2]

(b) An ellipse is defined parametrically by

$$x = 4\cos\theta, \qquad y = 3\sin\theta, \qquad 0 \leq \theta \leq 2\pi.$$

The arc length $s$ from $\theta = 0$ to $\theta = \pi/2$ is given by

$$s = \int_0^{\pi/2} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} ; d\theta.$$

Show that

$$s = \int_0^{\pi/2} 4\sqrt{1 - \frac{7}{16}\cos^2\theta} ; d\theta.$$

 [4]

(c) Use the binomial expansion from part (a) with a suitable substitution to expand the integrand $\sqrt{1 - \frac{7}{16}\cos^2\theta}$ up to and including the term in $\cos^4\theta$.

Hence, approximate $s$ by integrating term-by-term from $\theta = 0$ to $\theta = \pi/2$. You may use the standard results

$$\int_0^{\pi/2} \cos^2\theta , d\theta = \frac{\pi}{4}, \qquad \int_0^{\pi/2} \cos^4\theta , d\theta = \frac{3\pi}{16}.$$

Give your answer in terms of $\pi$, correct to 3 significant figures.  [8]

Question 6 [14 marks] — Pure Differential Equations

(a) Show that the differential equation

$$(3x^2y^2 + 2xy),dx + (2x^3y + x^2),dy = 0$$

is exact.  [3]

(b) Solve the exact equation to find its general solution.  [5]

(c) Determine whether the differential equation

$$(x + 2y + 1),dx + (2x + 4y - 3),dy = 0$$

is a non-homogeneous DE of linearly dependent or independent type. Justify your answer.

Hence, use the substitution $u = x + 2y$ to reduce the equation to a separable DE in $u$ and $y$, and solve it to find the general solution in implicit form.  [6]

Question 7 [14 marks] — Pure Maclaurin Series

(a) Find the Maclaurin series for $f(x) = \sin x$ using the definition (by computing successive derivatives at $x = 0$) up to and including the term in $x^5$.  [4]

(b) Using a standard expansion, write down the Maclaurin series for $g(x) = e^{x^2}$ up to and including the term in $x^6$.  [2]

(c) Hence, by multiplying the series from parts (a) and (b) term-by-term, find the Maclaurin series for

$$h(x) = e^{x^2}\sin x$$

up to and including the term in $x^5$.  [4]

(d) Use your series from part (c) to approximate the definite integral

$$\int_0^{0.5} e^{x^2}\sin x , dx$$

by integrating term-by-term up to $x^5$. Give your answer correct to 4 decimal places.

Explain why term-by-term integration of the Maclaurin series is necessary here, rather than finding an exact antiderivative.  [4]

Question 8 [14 marks] — Pure Ellipse

(a) Convert the general equation

$$4x^2 + 9y^2 - 16x + 18y - 11 = 0$$

into standard form by completing the square. Hence, determine the centre, vertices, foci, and eccentricity of the ellipse.  [6]

(b) Find the equations of the two tangent lines to this ellipse that are parallel to the line $4x + 3y = 5$.  [5]

(c) The point $P$ on the ellipse has $x$-coordinate $3$. Find the $y$-coordinates of $P$.

Calculate the sum of the distances from $P$ to the two foci. Verify that your answer equals $2a$, confirming the geometric definition of an ellipse.  [3]

END OF QUESTION PAPER

Answer Key

Part A

Q1 (a) $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \cdots = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + \cdots$

(b) $9(x^2 - 2x) + 4(y^2 + 4y) = 11$ $$9[(x-1)^2 - 1] + 4[(y+2)^2 - 4] = 11$$ $$9(x-1)^2 - 9 + 4(y+2)^2 - 16 = 11$$ $$9(x-1)^2 + 4(y+2)^2 = 36$$ $$\displaystyle \frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$$ Centre: $(1,-2)$, vertical orientation, vertices: $(1, -2\pm3) = (1,1), (1,-5)$

(c) $(y+1),dy = 2x,dx$ $$\frac{y^2}{2} + y = x^2 + C$$ $$y(0)=1 \Rightarrow \frac{1}{2} + 1 = C \Rightarrow C = \frac{3}{2}$$ $$y^2 + 2y = 2x^2 + 3$$ $$(y+1)^2 = 2x^2 + 4$$ $y = -1 + \sqrt{2x^2 + 4}$ (positive root since $y(0)=1 > -1$)

Q2 (a) $\ln(1+u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \dfrac{u^4}{4} + \cdots$, valid for $-1 < u \leq 1$. $\ln(1+3x^2) = 3x^2 - \dfrac{9x^4}{2} + 9x^6 - \cdots$, valid for $|x| \leq \dfrac{1}{\sqrt{3}}$.

(b) $a_n = \dfrac{5 - 2/n + 1/n^2}{2 + 3/n - 4/n^2} \to \dfrac{5}{2}$ as $n\to\infty$. Converges.

(c) $\dfrac{1}{(2k-1)(2k+1)} = \dfrac{1}{2}\left(\dfrac{1}{2k-1} - \dfrac{1}{2k+1}\right)$ $$\displaystyle \sum_{k=1}^n = \frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \frac{n}{2n+1}$$ Sum to infinity: $\displaystyle \lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2}$

(d) Vertical ellipse: $b = 5$, $c = 3$, $a^2 = b^2 - c^2 = 25 - 9 = 16$. $$\displaystyle \frac{(x-2)^2}{16} + \frac{(y+1)^2}{25} = 1$$

Part B

Q3 (a) $P(x) = 1$, $Q(x) = x$, $n = 2$.

(b) $v = y^{-1}$, $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$. $\dfrac{dy}{dx} = -y^2\dfrac{dv}{dx}$. Substitute: $-y^2\dfrac{dv}{dx} + y = xy^2$. Divide by $-y^2$: $\dfrac{dv}{dx} - y^{-1} = -x$. Since $y^{-1} = v$: $\dfrac{dv}{dx} - v = -x$.

(c) Integrating factor $I = e^{\int -1,dx} = e^{-x}$. $\dfrac{d}{dx}(v e^{-x}) = -xe^{-x}$. $v e^{-x} = \int -xe^{-x},dx = xe^{-x} + e^{-x} + C$. $v = x + 1 + Ce^{x}$. $y = \dfrac{1}{x + 1 + Ce^{x}}$.

(d) $y(0) = 1 \Rightarrow 1 = \dfrac{1}{1 + C} \Rightarrow C = 0$. $y = \dfrac{1}{x+1}$.

From DE: $y'(0) = x(0)\cdot y(0)^2 - y(0) = 0 - 1 = -1$. Differentiate DE: $y'' + y' = y^2 + 2xyy'$ $\Rightarrow y''(0) = y(0)^2 + 0 - y'(0) = 1 + 1 = 2$. Differentiate again: $y''' + y'' = 2yy' + 2yy' + 2x(y'^2 + yy'') = 4yy' + 2x(y'^2 + yy'')$ $\Rightarrow y'''(0) = 4(1)(-1) + 0 - y''(0) = -4 - 2 = -6$.

Maclaurin series: $y(x) = 1 - x + x^2 - x^3 + \cdots$

Q4 (a) $4y,dy = -9x,dx$. $$\int 4y,dy = -\int 9x,dx$$ $$2y^2 = -\dfrac{9}{2}x^2 + C$$ $9x^2 + 4y^2 = 2C$ (general solution).

(b) At $(2,0)$: $9(4) + 4(0) = 36 = 2C \Rightarrow C = 18$. $9x^2 + 4y^2 = 36 \Rightarrow \dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.

(c) Ellipse, vertical orientation ($b > a$ where $b=3$, $a=2$). Centre: $(0,0)$. Vertices: $(0, \pm 3)$. $c^2 = b^2 - a^2 = 9 - 4 = 5$, $c = \sqrt{5}$. Foci: $(0, \pm\sqrt{5})$.

(d) Verify $\left(\dfrac{4}{3}, \sqrt{5}\right)$ lies on $9x^2+4y^2=36$: $9\left(\dfrac{16}{9}\right) + 4(5) = 16 + 20 = 36$ ✓.

$\dfrac{dy}{dx}\bigg|_{(4/3, \sqrt{5})} = -\dfrac{9(4/3)}{4\sqrt{5}} = -\dfrac{12}{4\sqrt{5}} = -\dfrac{3}{\sqrt{5}}$.

Tangent: $y - \sqrt{5} = -\dfrac{3}{\sqrt{5}}\left(x - \dfrac{4}{3}\right)$ $$\Rightarrow y = -\dfrac{3}{\sqrt{5}}x + \dfrac{4}{\sqrt{5}} + \sqrt{5}$$ $\Rightarrow y = -\frac{3}{\sqrt{5}}x + \frac{9}{\sqrt{5}}$.

Q5 (a) $(1+u)^{-1/2} = 1 - \dfrac{1}{2}u + \dfrac{3}{8}u^2 - \dfrac{5}{16}u^3 + \cdots$, valid for $|u| < 1$.

(b) $\dfrac{dx}{d\theta} = -4\sin\theta$, $\dfrac{dy}{d\theta} = 3\cos\theta$. $$\left(\dfrac{dx}{d\theta}\right)^2 + \left(\dfrac{dy}{d\theta}\right)^2 = 16\sin^2\theta + 9\cos^2\theta$$ $= 16(1-\cos^2\theta) + 9\cos^2\theta = 16 - 7\cos^2\theta$. $\sqrt{16 - 7\cos^2\theta} = 4\sqrt{1 - \frac{7}{16}\cos^2\theta}$. Hence $s = \displaystyle\int_0^{\pi/2} 4\sqrt{1 - \frac{7}{16}\cos^2\theta},d\theta$.

(c) Substitute $u = -\frac{7}{16}\cos^2\theta$ into $(1+u)^{1/2}$: $$(1+u)^{1/2} = 1 + \frac{1}{2}u - \frac{1}{8}u^2 + \cdots$$ $\sqrt{1 - \frac{7}{16}c^2} = 1 - \frac{7}{32}c^2 - \frac{49}{2048}c^4 + \cdots$, where $c = \cos\theta$.

$$s \approx 4\int_0^{\pi/2} \left(1 - \frac{7}{32}\cos^2\theta - \frac{49}{2048}\cos^4\theta\right) d\theta$$

$$\int_0^{\pi/2} 1,d\theta = \dfrac{\pi}{2}$$ $$\int_0^{\pi/2} \cos^2\theta,d\theta = \dfrac{\pi}{4}$$ $$\int_0^{\pi/2} \cos^4\theta,d\theta = \dfrac{3\pi}{16}$$

$$s \approx 4\left[\dfrac{\pi}{2} - \dfrac{7}{32}\cdot\dfrac{\pi}{4} - \dfrac{49}{2048}\cdot\dfrac{3\pi}{16}\right]$$ $$= 4\pi\left[\dfrac{1}{2} - \dfrac{7}{128} - \dfrac{147}{32768}\right]$$ $$= 4\pi\left[\dfrac{16384 - 1792 - 147}{32768}\right]$$ $$= 4\pi\left[\dfrac{14445}{32768}\right]$$ $$= \dfrac{14445\pi}{8192}$$ $\approx 5.54$ (3 s.f.)

Q6 (a) $M = 3x^2y^2 + 2xy$, $N = 2x^3y + x^2$. $\dfrac{\partial M}{\partial y} = 6x^2y + 2x$, $\dfrac{\partial N}{\partial x} = 6x^2y + 2x$. $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ ✓. Exact.

(b) $F = \int M,dx = \int (3x^2y^2 + 2xy),dx = x^3y^2 + x^2y + g(y)$. $\dfrac{\partial F}{\partial y} = 2x^3y + x^2 + g'(y) = N = 2x^3y + x^2$. $\Rightarrow g'(y) = 0 \Rightarrow g(y) = \text{constant}$. General solution: $x^3y^2 + x^2y = C$.

(c) $(x + 2y + 1)dx + (2x + 4y - 3)dy = 0$. $a_1 = 1$, $b_1 = 2$, $a_2 = 2$, $b_2 = 4$. $a_1b_2 - a_2b_1 = 1(4) - 2(2) = 4 - 4 = 0$. Hence linearly dependent.

Let $u = x + 2y$. Then $du = dx + 2,dy \Rightarrow dx = du - 2,dy$. Substituting: $(u + 1)(du - 2,dy) + (2u - 3)dy = 0$ $$(u+1)du - 2(u+1)dy + (2u-3)dy = 0$$ $$(u+1)du + (-2u - 2 + 2u - 3)dy = 0$$ $$(u+1)du - 5,dy = 0$$ $\dfrac{du}{dy} = \dfrac{5}{u+1}$ (separable in $u$ and $y$)

$$(u+1)du = 5,dy$$ $$\int (u+1)du = \int 5,dy$$ $$\dfrac{u^2}{2} + u = 5y + C$$ $$\dfrac{(x+2y)^2}{2} + (x+2y) = 5y + C$$

General solution (implicit): $(x+2y)^2 + 2(x+2y) = 10y + K$, where $K = 2C$.

Q7 (a) $f(x) = \sin x$, $f(0) = 0$. $f'(x) = \cos x$, $f'(0) = 1$. $f''(x) = -\sin x$, $f''(0) = 0$. $f'''(x) = -\cos x$, $f'''(0) = -1$. $f^{(4)}(x) = \sin x$, $f^{(4)}(0) = 0$. $f^{(5)}(x) = \cos x$, $f^{(5)}(0) = 1$.

$$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + \cdots = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} + \cdots$$

(b) $e^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \cdots = 1 + x^2 + \dfrac{x^4}{2} + \dfrac{x^6}{6} + \cdots$

(c) $h(x) = e^{x^2}\sin x = \left(1 + x^2 + \dfrac{x^4}{2} + \cdots\right)\left(x - \dfrac{x^3}{6} + \dfrac{x^5}{120} + \cdots\right)$

$x$ term: $1\cdot x = x$ $x^3$ term: $1(-\frac{x^3}{6}) + x^2(x) = -\frac{x^3}{6} + x^3 = \frac{5}{6}x^3$ $x^5$ term: $1(\frac{x^5}{120}) + x^2(-\frac{x^3}{6}) + \frac{x^4}{2}(x)$ $$= \frac{x^5}{120} - \frac{x^5}{6} + \frac{x^5}{2}$$ $$= \left(\frac{1}{120} - \frac{20}{120} + \frac{60}{120}\right)x^5 = \frac{41}{120}x^5$$

$$h(x) = x + \dfrac{5}{6}x^3 + \dfrac{41}{120}x^5 + \cdots$$

(d) $\displaystyle\int_0^{0.5} h(x),dx \approx \int_0^{0.5} \left(x + \frac{5}{6}x^3 + \frac{41}{120}x^5\right) dx$ $$= \left[\dfrac{x^2}{2} + \dfrac{5}{24}x^4 + \dfrac{41}{720}x^6\right]_0^{0.5}$$ $$= \dfrac{(0.5)^2}{2} + \dfrac{5}{24}(0.5)^4 + \dfrac{41}{720}(0.5)^6$$ $$= \dfrac{0.25}{2} + \dfrac{5}{24}(0.0625) + \dfrac{41}{720}(0.015625)$$ $$= 0.125 + 0.0130208 + 0.0008898$$ $= 0.1389$ (4 d.p.)

The function $e^{x^2}\sin x$ has no elementary antiderivative — its integral cannot be expressed as a finite combination of standard functions. Maclaurin series term-by-term integration is the standard method for approximating such integrals.

Q8 (a) $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ $$4(x^2 - 4x) + 9(y^2 + 2y) = 11$$ $$4[(x-2)^2 - 4] + 9[(y+1)^2 - 1] = 11$$ $$4(x-2)^2 - 16 + 9(y+1)^2 - 9 = 11$$ $$4(x-2)^2 + 9(y+1)^2 = 36$$ $$\dfrac{(x-2)^2}{9} + \dfrac{(y+1)^2}{4} = 1$$

Centre: $(2, -1)$. Horizontal ellipse. $a = 3$, $b = 2$. $c^2 = a^2 - b^2 = 9 - 4 = 5$, $c = \sqrt{5}$. Vertices: $(2\pm3, -1) = (5, -1)$, $(-1, -1)$. Foci: $(2\pm\sqrt{5}, -1)$. Eccentricity: $e = \dfrac{c}{a} = \dfrac{\sqrt{5}}{3}$.

(b) Line $4x+3y=5$ has gradient $m = -\dfrac{4}{3}$. Translate ellipse: $X = x-2$, $Y = y+1$ gives $\dfrac{X^2}{9} + \dfrac{Y^2}{4} = 1$. Tangents with gradient $m$: $Y = mX \pm \sqrt{a^2m^2 + b^2}$. $a^2m^2 + b^2 = 9\left(\dfrac{16}{9}\right) + 4 = 16 + 4 = 20$. $\sqrt{a^2m^2 + b^2} = \sqrt{20} = 2\sqrt{5}$.

$Y = -\dfrac{4}{3}X \pm 2\sqrt{5}$. In original coordinates: $$y+1 = -\dfrac{4}{3}(x-2) \pm 2\sqrt{5}$$ $$y = -\dfrac{4}{3}x + \dfrac{8}{3} - 1 \pm 2\sqrt{5}$$ $y = -\dfrac{4}{3}x + \dfrac{5}{3} \pm 2\sqrt{5}$.

The two tangent lines: $y = -\dfrac{4}{3}x + \dfrac{5}{3} + 2\sqrt{5}$ and $y = -\dfrac{4}{3}x + \dfrac{5}{3} - 2\sqrt{5}$.

(c) At $x = 3$: $\dfrac{(3-2)^2}{9} + \dfrac{(y+1)^2}{4} = 1$ $$\dfrac{1}{9} + \dfrac{(y+1)^2}{4} = 1$$ $$\dfrac{(y+1)^2}{4} = \dfrac{8}{9}$$ $$(y+1)^2 = \dfrac{32}{9}$$ $$y+1 = \pm\dfrac{4\sqrt{2}}{3}$$ $y = -1 \pm \dfrac{4\sqrt{2}}{3}$.

Take $P = \left(3, -1 + \dfrac{4\sqrt{2}}{3}\right)$. Foci: $F_1 = (2-\sqrt{5}, -1)$, $F_2 = (2+\sqrt{5}, -1)$.

$PF_1 = \sqrt{(3-(2-\sqrt{5}))^2 + \left(\dfrac{4\sqrt{2}}{3}\right)^2} = \sqrt{(1+\sqrt{5})^2 + \dfrac{32}{9}} = \sqrt{1 + 2\sqrt{5} + 5 + \dfrac{32}{9}} = \sqrt{6 + 2\sqrt{5} + \dfrac{32}{9}} = \sqrt{\dfrac{86}{9} + \dfrac{18\sqrt{5}}{9}} = \dfrac{1}{3}\sqrt{86 + 18\sqrt{5}}$

$PF_2 = \sqrt{(3-(2+\sqrt{5}))^2 + \left(\dfrac{4\sqrt{2}}{3}\right)^2} = \sqrt{(1-\sqrt{5})^2 + \dfrac{32}{9}} = \sqrt{1 - 2\sqrt{5} + 5 + \dfrac{32}{9}} = \sqrt{6 - 2\sqrt{5} + \dfrac{32}{9}} = \sqrt{\dfrac{86}{9} - \dfrac{18\sqrt{5}}{9}} = \dfrac{1}{3}\sqrt{86 - 18\sqrt{5}}$

Since $86 + 18\sqrt{5} = (9+\sqrt{5})^2$ and $86 - 18\sqrt{5} = (9-\sqrt{5})^2$: $PF_1 + PF_2 = \dfrac{1}{3}(9+\sqrt{5} + 9-\sqrt{5}) = \dfrac{1}{3}(18) = 6 = 2a$ ✓

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