Question
Solve the equation \(4 \sin \theta+\tan \theta=0\) for \(0^{\circ}<\theta<180^{\circ}\).
▶️Answer/Explanation
To solve \(4 \sin \theta + \tan \theta = 0\) for \(0^\circ < \theta < 180^\circ\), start by rewriting \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\), so the equation becomes:
\[ 4 \sin \theta + \frac{\sin \theta}{\cos \theta} = 0 \]
Factor out \(\sin \theta\):
\[ \sin \theta \left( 4 + \frac{1}{\cos \theta} \right) = 0 \]
This gives two possibilities: \(\sin \theta = 0\) or \(4 + \frac{1}{\cos \theta} = 0\).
If \(\sin \theta = 0\), then \(\theta = 0^\circ\) or \(\theta = 180^\circ\), but since the range is \(0^\circ < \theta < 180^\circ\), neither works.
If \(4 + \frac{1}{\cos \theta} = 0\), solve for \(\cos \theta\):
\[ \frac{1}{\cos \theta} = -4 \]
\[ \cos \theta = -\frac{1}{4} \]
Since \(\theta\) is between \(0^\circ\) and \(180^\circ\), and \(\cos \theta = -\frac{1}{4}\) is negative, \(\theta\) must be in the second quadrant. Using a calculator, \(\cos^{-1}\left(\frac{1}{4}\right) \approx 75.52^\circ\), so:
\[ \theta = 180^\circ – 75.52^\circ = 104.48^\circ \]
Answer is:
\[ \theta \approx 104.5^\circ \]
Question
(a) Find the first three terms in the expansion, in ascending powers of \(x\), of \((2+3 x)^4\).
(b) Find the first three terms in the expansion, in ascending powers of \(x\), of \((1-2 x)^5\).
(c) Hence find the coefficient of \(x^2\) in the expansion of \((2+3 x)^4(1-2 x)^5\).
▶️Answer/Explanation
(a) First three terms of \((2 + 3x)^4\)
Use the binomial theorem: \((a + b)^n = a^n + n a^{n-1} b + \frac{n(n-1)}{2} a^{n-2} b^2 + \cdots\), where \(a = 2\), \(b = 3x\), and \(n = 4\).
Term 1: \(2^4 = 16\)
Term 2: \(4 \cdot 2^{4-1} \cdot (3x) = 4 \cdot 8 \cdot 3x = 96x\)
Term 3: \(\frac{4 \cdot 3}{2} \cdot 2^{4-2} \cdot (3x)^2 = 6 \cdot 4 \cdot 9x^2 = 216x^2\)
So, the first three terms are:
\[ 16 + 96x + 216x^2 \]
(b) First three terms of \((1 – 2x)^5\)
Here, \(a = 1\), \(b = -2x\), and \(n = 5\).
Term 1: \(1^5 = 1\)
Term 2: \(5 \cdot 1^{5-1} \cdot (-2x) = 5 \cdot 1 \cdot (-2x) = -10x\)
Term 3: \(\frac{5 \cdot 4}{2} \cdot 1^{5-2} \cdot (-2x)^2 = 10 \cdot 1 \cdot 4x^2 = 40x^2\)
So, the first three terms are:
\[ 1 – 10x + 40x^2 \]
(c) Coefficient of \(x^2\) in \((2 + 3x)^4 (1 – 2x)^5\)
To find the \(x^2\) coefficient, multiply the expansions and collect terms that give \(x^2\):
\(x^0\) from \((2 + 3x)^4\) × \(x^2\) from \((1 – 2x)^5\): \(16 \cdot 40x^2 = 640x^2\)
\(x^1\) from \((2 + 3x)^4\) × \(x^1\) from \((1 – 2x)^5\): \(96x \cdot (-10x) = -960x^2\)
\(x^2\) from \((2 + 3x)^4\) × \(x^0\) from \((1 – 2x)^5\): \(216x^2 \cdot 1 = 216x^2\)
Add them up:
\[ 640 + (-960) + 216 = 640 – 960 + 216 = -104 \]
So, the coefficient of \(x^2\) is:
\[ -104 \]
Final Answers:
(a) \(16 + 96x + 216x^2\)
(b) \(1 – 10x + 40x^2\)
(c) \(-104\)
Question
The diagram shows graphs with equations \(y=\mathrm{f}(x)\) and \(y=\mathrm{g}(x)\).
Describe fully a sequence of two transformations which transforms the graph of \(y=\mathrm{f}(x)\) to \(y=\mathrm{g}(x)\).
▶️Answer/Explanation
Ans:
\(\{\) Stretch \(\}\{\) factor 2\(\}\) in \(y\)-direction \(\}\)
\(\{\) Translation \(\}\left(\begin{array}{c}\{-6\} \\ \{0\}\end{array}\right)\)
Question
The diagram shows a sector \(A B C\) of a circle with centre \(A\) and radius \(8 \mathrm{~cm}\). The area of the sector is \(\frac{16}{3} \pi \mathrm{cm}^2\). The point \(D\) lies on the \(\operatorname{arc} B C\).
Find the perimeter of the segment \(B C D\).
▶️Answer/Explanation
Ans:
$
\frac{1}{2} \times 8^2 \times \theta=\frac{16 \pi}{3} \Rightarrow \theta=\frac{\pi}{6}
$
Arc length \(=8 \times\) their \(\frac{\pi}{6}[=4.1887 \ldots]\)
$
[B C=] 2 \times 8 \sin \left(\frac{1}{2} \times \text { their } \frac{\pi}{6}\right)[=4.1411 \ldots]
$
$
\text { Perimeter }=8.33
$
Alternative methods for Question 4: 2nd M1 mark (use normal scheme for the other marks)
ALT \(1 B C^2=8^2+8^2-2 \times 8 \times 8 \cos \left(\right.\) their \(\left.\frac{\pi}{6}\right)[\Rightarrow B C=4.14 \ldots]\)
ALT \(2 B C^2=(8-4 \sqrt{3})^2+4^2[\Rightarrow B C=4.14 \ldots]\)
ALT \(3 \frac{B C}{\sin \left(\frac{\pi}{6}\right)}=\frac{8}{\sin \left(\frac{5 \pi}{12}\right)}[\Rightarrow B C=4.14 \ldots]\)
Question
The line with equation \(y=k x-k\), where \(k\) is a positive constant, is a tangent to the curve with equation \(y=-\frac{1}{2 x}\).
Find, in either order, the value of \(k\) and the coordinates of the point where the tangent meets the curve.
▶️Answer/Explanation
Step 1: Set the equations equal (point of contact)
Since the line touches the curve:
\[ kx – k = -\frac{1}{2x} \]
Multiply through by \(2x\) to clear the fraction:
\[ 2x(kx – k) = -1 \]
\[ 2kx^2 – 2kx = -1 \]
Rearrange into standard form:
\[ 2kx^2 – 2kx + 1 = 0 \]
This is a quadratic in \(x\). For tangency, it has exactly one solution, so the discriminant must be zero.
Step 2: Use the discriminant
For \(ax^2 + bx + c = 0\), the discriminant is \(\Delta = b^2 – 4ac\). Here, \(a = 2k\), \(b = -2k\), \(c = 1\):
\[ \Delta = (-2k)^2 – 4 \cdot 2k \cdot 1 = 4k^2 – 8k \]
Set \(\Delta = 0\):
\[ 4k^2 – 8k = 0 \]
\[ 4k(k – 2) = 0 \]
Solutions: \(k = 0\) or \(k = 2\). Since \(k\) is positive, \(k = 2\).
Step 3: Find the point of contact
Substitute \(k = 2\) into the quadratic:
\[ 2 \cdot 2 \cdot x^2 – 2 \cdot 2 \cdot x + 1 = 0 \]
\[ 4x^2 – 4x + 1 = 0 \]
Discriminant: \(16 – 16 = 0\), confirming one solution. Solve:
\[ (2x – 1)^2 = 0 \]
\[ 2x – 1 = 0 \]
\[ x = \frac{1}{2} \]
Now, use the curve’s equation:
\[ y = -\frac{1}{2 \cdot \frac{1}{2}} = -\frac{1}{1} = -1 \]
So, the point is \(\left(\frac{1}{2}, -1\right)\).
Step 4: Verify tangency (optional)
Slope of the curve: \(y = -\frac{1}{2}x^{-1}\), so \(\frac{dy}{dx} = \frac{1}{2}x^{-2} = \frac{1}{2x^2}\). At \(x = \frac{1}{2}\):
\[ \frac{1}{2 \cdot \left(\frac{1}{2}\right)^2} = \frac{1}{2 \cdot \frac{1}{4}} = 2 \]
Slope of the line: \(y = 2x – 2\), slope = 2. They match, confirming tangency.
Final Answer:
Value of \(k\): \(2\)
Point of tangency: \(\left(\frac{1}{2}, -1\right)\)
Question
The first three terms of an arithmetic progression are \(\frac{p^2}{6}, 2 p-6\) and \(p\).
(a) Given that the common difference of the progression is not zero, find the value of \(p\).
(b) Using this value, find the sum to infinity of the geometric progression with first two terms \(\frac{p^2}{6}\) and \(2 p-6\).
▶️Answer/Explanation
(a) Find the value of \(p\)
In an arithmetic progression, the difference between consecutive terms is constant. Given the first three terms \(\frac{p^2}{6}\), \(2p – 6\), and \(p\), the common difference \(d\) is:
\(d = (2p – 6) – \frac{p^2}{6}\) (second term minus first)
\(d = p – (2p – 6)\) (third term minus second)
Since \(d \neq 0\), set the two expressions for \(d\) equal:
\[ (2p – 6) – \frac{p^2}{6} = p – (2p – 6) \]
Simplify:
Left side: \(2p – 6 – \frac{p^2}{6}\)
Right side: \(p – 2p + 6 = -p + 6\)
So:
\[ 2p – 6 – \frac{p^2}{6} = -p + 6 \]
Multiply through by 6 to clear the fraction:
\[ 6(2p – 6) – p^2 = 6(-p + 6) \]
\[ 12p – 36 – p^2 = -6p + 36 \]
Move all terms to one side:
\[ -p^2 + 12p – 36 + 6p – 36 = 0 \]
\[ -p^2 + 18p – 72 = 0 \]
Multiply by \(-1\):
\[ p^2 – 18p + 72 = 0 \]
Factorize:
\[ (p – 6)(p – 12) = 0 \]
Solutions: \(p = 6\) or \(p = 12\).
Check \(d \neq 0\):
If \(p = 6\): Terms are \(\frac{6^2}{6} = 6\), \(2 \cdot 6 – 6 = 6\), \(6\). Differences: \(6 – 6 = 0\), \(6 – 6 = 0\). \(d = 0\), so discard.
If \(p = 12\): Terms are \(\frac{12^2}{6} = 24\), \(2 \cdot 12 – 6 = 18\), \(12\). Differences: \(18 – 24 = -6\), \(12 – 18 = -6\). \(d = -6 \neq 0\), so it works.
Thus, \(p = 12\).
(b) Sum to infinity of the geometric progression
Use \(p = 12\). First two terms are \(\frac{p^2}{6} = 24\) and \(2p – 6 = 18\). For a geometric progression, the common ratio \(r = \frac{\text{second term}}{\text{first term}} = \frac{18}{24} = \frac{3}{4}\).
Sum to infinity: \(S = \frac{a}{1 – r}\), where \(a = 24\) and \(r = \frac{3}{4}\):
\[ S = \frac{24}{1 – \frac{3}{4}} = \frac{24}{\frac{1}{4}} = 24 \cdot 4 = 96 \]
Final Answer:
(a) \(p = 12\)
(b) \(96\)
Question
(b) Topic 1.5 – Trigonometry
(c) Topic 1.3 – Coordinate geometry
A curve has equation \(y=2+3 \sin \frac{1}{2} x\) for \(0 \leqslant x \leqslant 4 \pi\).
(a) State greatest and least values of \(y\).
(b) Sketch the curve.
(c) State the number of solutions of the equation
$
2+3 \sin \frac{1}{2} x=5-2 x
$
for \(0 \leqslant x \leqslant 4 \pi\).
▶️Answer/Explanation
Ans:
a.[Greatest =] 5
[Least =] ‒1
b.
(c) 1
Question
Topic 1.2 – Functions
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined as follows, where \(a\) and \(b\) are constants.
$
\begin{aligned}
& \mathrm{f}(x)=1+\frac{2 a}{x-a} \text { for } x>a \\
& \mathrm{~g}(x)=b x-2 \text { for } x \in \mathbb{R}
\end{aligned}
$
(a) Given that \(\mathrm{f}(7)=\frac{5}{2}\) and \(\operatorname{gf}(5)=4\), find the values of \(a\) and \(b\).
For the rest of this question, you should use the value of \(a\) which you found in (a).
(b) Find the domain of \(\mathrm{f}^{-1}\).
(c) Find an expression for \(\mathrm{f}^{-1}(x)\).
▶️Answer/Explanation
(a) Find \(a\) and \(b\)
1. Use \(\mathrm{f}(7) = \frac{5}{2}\)
\(\mathrm{f}(x) = 1 + \frac{2a}{x – a}\), so:
\(\mathrm{f}(7) = 1 + \frac{2a}{7 – a} = \frac{5}{2}\)
Subtract 1:
\(\frac{2a}{7 – a} = \frac{5}{2} – 1 = \frac{3}{2}\)
Cross-multiply:
\(2a \cdot 2 = 3 (7 – a)\)
\(4a = 21 – 3a\)
\(7a = 21\)
\(a = 3\)
2. Use \(\mathrm{gf}(5) = 4\)
First, find \(\mathrm{f}(5)\) with \(a = 3\):
\(\mathrm{f}(5) = 1 + \frac{2 \cdot 3}{5 – 3} = 1 + \frac{6}{2} = 1 + 3 = 4\)
Now, \(\mathrm{g}(x) = bx – 2\), so:
\(\mathrm{gf}(5) = \mathrm{g}(\mathrm{f}(5)) = \mathrm{g}(4) = b \cdot 4 – 2 = 4\)
Solve:
\(4b – 2 = 4\)
\(4b = 6\)
\(b = \frac{6}{4} = \frac{3}{2}\)
So, \(a = 3\), \(b = \frac{3}{2}\).
(b) Domain of \(\mathrm{f}^{-1}\)
Use \(a = 3\). The inverse \(\mathrm{f}^{-1}\) swaps the domain and range of \(\mathrm{f}\).
Domain of \(\mathrm{f}\): \(x > 3\).
Range of \(\mathrm{f}\): Find \(y = \mathrm{f}(x) = 1 + \frac{6}{x – 3}\). As \(x \to 3^+\), \(y \to \infty\); as \(x \to \infty\), \(y \to 1^+\). So, \(y > 1\).
Thus, domain of \(\mathrm{f}^{-1}\): \(x > 1\).
(c) Expression for \(\mathrm{f}^{-1}(x)\)
Start with \(y = \mathrm{f}(x) = 1 + \frac{6}{x – 3}\) (using \(a = 3\)). Solve for \(x\):
\(y – 1 = \frac{6}{x – 3}\)
\(x – 3 = \frac{6}{y – 1}\)
\(x = 3 + \frac{6}{y – 1}\)
So, \(\mathrm{f}^{-1}(x) = 3 + \frac{6}{x – 1}\).
Final Answer:
(a) \(a = 3\), \(b = \frac{3}{2}\)
(b) \(x > 1\)
(c) \(\mathrm{f}^{-1}(x) = 3 + \frac{6}{x – 1}\)
Question
Topic 1.7 – Differentiation
Water is poured into a tank at a constant rate of \(500 \mathrm{~cm}^3\) per second. The depth of water in the tank, \(t\) seconds after filling starts, is \(h \mathrm{~cm}\). When the depth of water in the tank is \(h \mathrm{~cm}\), the volume, \(V \mathrm{~cm}^3\), of water in the tank is given by the formula \(V=\frac{4}{3}(25+h)^3-\frac{62500}{3}\).
(a) Find the rate at which \(h\) is increasing at the instant when \(h=10 \mathrm{~cm}\).
(b) At another instant, the rate at which h is increasing is 0.075 cm per second.
Find the value of V at this instant.
▶️Answer/Explanation
(a) Rate at which \(h\) is increasing when \(h = 10 \, \text{cm}\)
Water is poured at \(500 \, \text{cm}^3/\text{s}\), so \(\frac{dV}{dt} = 500\). The volume is given by:
\[ V = \frac{4}{3}(25 + h)^3 – \frac{62500}{3} \]
To find \(\frac{dh}{dt}\), use the chain rule: \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\).
Differentiate \(V\) with respect to \(h\):
\(\frac{dV}{dh} = \frac{d}{dh} \left[ \frac{4}{3}(25 + h)^3 – \frac{62500}{3} \right] = \frac{4}{3} \cdot 3 (25 + h)^2 = 4 (25 + h)^2\)
At \(h = 10\):
\(\frac{dV}{dh} = 4 (25 + 10)^2 = 4 \cdot 35^2 = 4 \cdot 1225 = 4900\)
Solve for \(\frac{dh}{dt}\):
\(500 = 4900 \cdot \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{500}{4900} = \frac{5}{49} \, \text{cm/s}\)
So, the rate is \(\frac{5}{49} \, \text{cm/s}\).
(b) Find \(V\) when \(\frac{dh}{dt} = 0.075 \, \text{cm/s}\)
Using \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\):
\(\frac{dV}{dt} = 500\)
\(\frac{dV}{dh} = 4 (25 + h)^2\)
\(\frac{dh}{dt} = 0.075\)
So:
\[ 500 = 4 (25 + h)^2 \cdot 0.075 \]
\[ 500 = 0.3 (25 + h)^2 \]
\[ (25 + h)^2 = \frac{500}{0.3} = \frac{5000}{3} \approx 1666.67 \]
\[ 25 + h = \sqrt{\frac{5000}{3}} \approx 40.825 \]
\[ h \approx 40.825 – 25 = 15.825 \, \text{cm} \]
Now, calculate \(V\):
\[ V = \frac{4}{3} (25 + 15.825)^3 – \frac{62500}{3} \]
\[ V = \frac{4}{3} (40.825)^3 – \frac{62500}{3} \]
Approximate: \((40.825)^3 \approx 68016.7\) (using a calculator), so:
\[ V = \frac{4}{3} \cdot 68016.7 – \frac{62500}{3} \approx 90688.93 – 20833.33 \approx 69855.6 \]
For simplicity, solve exactly if needed, but numerically:
\(V \approx 69856 \, \text{cm}^3\).
Final Answer:
(a) \(\frac{5}{49} \, \text{cm/s}\)
(b) \(69856 \, \text{cm}^3\) (approximate)
Question
(b) Topic 1.7 – Differentiation
The diagram shows part of the curve with equation \(y=\frac{4}{(2 x-1)^2}\) and parts of the lines \(x=1\) and \(y=1\). The curve passes through the points \(A(1,4)\) and \(B,\left(\frac{3}{2}, 1\right)\).
(a) Find the exact volume generated when the shaded region is rotated through \(360^{\circ}\) about the \(x\)-axis.
(b) A triangle is formed from the tangent to the curve at \(B\), the normal to the curve at \(B\) and the \(x\)-axis.
Find the area of this triangle.
▶️Answer/Explanation
Ans:
\([\pi] \int \frac{16}{(2 x-1)^4}[\mathrm{~d} x]=[\pi] \int 16(2 x-1)^{-4}[\mathrm{~d} x]=[\pi]\left(-\frac{16}{3 \times 2 \times(2 x-1)^3}\right)\)
\([\pi]\left(-\frac{16}{3 \times 2 \times(2 x-1)^3}\right)\)
\([\pi]\left(-\frac{16}{6 \times 8}+\frac{16}{6 \times 1}\right)\left[=[\pi] \frac{112}{48}=[\pi] \frac{7}{3}\right]\)
Volume of cylinder \(\left[=\pi \times 1^2 \times \frac{1}{2}\right]=\frac{1}{2} \pi\) OR \([\pi] \int_1^{1.5} 1[\mathrm{~d} x]=\frac{1}{2} \pi\)
Volume of revolution \(\left[=\frac{7}{3} \pi-\frac{1}{2} \pi\right]=\frac{11}{6} \pi\)
b.$
\left[\frac{\mathrm{d} y}{\mathrm{~d} x}=\right]\left\{-8(2 x-1)^{-3}\right\}\{\times 2\}
$
At \(B\) gradient \(=-2\)
Eqn of tangent \(y-1=\) their \(“-2 “\left(x-\frac{3}{2}\right)\) OR Eqn of normal \(y-1=\) their \(” \frac{1}{2} “\left(x-\frac{3}{2}\right)\)
Tangent crosses \(x\)-axis at 2 or normal crosses \(x\)-axis at \(-\frac{1}{2}\)
$
\text { Area }=\frac{5}{4}
$
Question
(b) Topic 1.7 – Differentiation
(c) Topic 1.8 – Integration
(d) Topic 1.7 – Differentiation
The equation of a curve is such that \(\frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^2-30 x+6 a\), where \(a\) is a positive constant. The curve has a stationary point at \((a,-15)\).
(a) Find the value of \(a\).
(b) Determine the nature of this stationary point.
(c) Find the equation of the curve.
(d) Find the coordinates of any other stationary points on the curve.
▶️Answer/Explanation
(a) Find the value of \(a\)
The curve has a stationary point at \((a, -15)\), meaning \(\frac{dy}{dx} = 0\) at \(x = a\). Given:
\[ \frac{dy}{dx} = 6x^2 – 30x + 6a \]
Set it to zero at \(x = a\):
\[ 6a^2 – 30a + 6a = 0 \]
\[ 6a^2 – 24a = 0 \]
\[ 6a(a – 4) = 0 \]
Since \(a > 0\), \(a = 4\).
So, \(a = 4\).
(b) Nature of the stationary point
Use the second derivative to determine the type. First derivative:
\[ \frac{dy}{dx} = 6x^2 – 30x + 24 \] (with \(a = 4\))
Second derivative:
\[ \frac{d^2y}{dx^2} = 12x – 30 \]
At \(x = a = 4\):
\[ \frac{d^2y}{dx^2} = 12 \cdot 4 – 30 = 48 – 30 = 18 \]
Since \(18 > 0\), it’s a minimum.
So, the stationary point is a minimum.
(c) Equation of the curve
Integrate \(\frac{dy}{dx} = 6x^2 – 30x + 24\):
\[ y = \int (6x^2 – 30x + 24) \, dx = 2x^3 – 15x^2 + 24x + c \]
Use the point \((4, -15)\):
\[ -15 = 2(4)^3 – 15(4)^2 + 24(4) + c \]
\[ -15 = 2 \cdot 64 – 15 \cdot 16 + 96 + c \]
\[ -15 = 128 – 240 + 96 + c \]
\[ -15 = -16 + c \]
\[ c = 1 \]
So, the equation is:
\[ y = 2x^3 – 15x^2 + 24x + 1 \]
(d) Other stationary points
Set \(\frac{dy}{dx} = 6x^2 – 30x + 24 = 0\):
\[ 6(x^2 – 5x + 4) = 0 \]
\[ x^2 – 5x + 4 = 0 \]
\[ (x – 1)(x – 4) = 0 \]
\[ x = 1 \text{ or } x = 4 \]
\(x = 4\) is the given point.
At \(x = 1\):
\(y = 2(1)^3 – 15(1)^2 + 24(1) + 1 = 2 – 15 + 24 + 1 = 12\)
So, the other stationary point is \((1, 12)\).
Final Answer:
(a) \(a = 4\)
(b) Minimum
(c) \(y = 2x^3 – 15x^2 + 24x + 1\)
(d) \((1, 12)\)
Question
The diagram shows a circle \(P\) with centre \((0,2)\) and radius 10 and the tangent to the circle at the point \(A\) with coordinates \((6,10)\). It also shows a second circle \(Q\) with centre at the point where this tangent meets the \(y\)-axis and with radius \(\frac{5}{2} \sqrt{5}\).
(a) Write down the equation of circle \(P\).
(b) Find the equation of the tangent to the circle P at A.
(c) Find the equation of circle Q and hence verify that the y-coordinates of both of the points of intersection of the two circles are 11.
(d) Find the coordinates of the points of intersection of the tangent and circle Q, giving the answers in surd form.
▶️Answer/Explanation
Ans:
a. \(x^2+(y-2)^2=100\)
b.Gradient of radius \(=\left[\frac{10-2}{6-0}=\right] \frac{4}{3}\) or gradient of tangent \(=\frac{-3}{4}\)
Equation of tangent is \(y-10=-\frac{3}{4}(x-6) \quad\left[\Rightarrow y=-\frac{3}{4} x+\frac{29}{2}\right]\)
Equation of tangent is \(y-10=-\frac{3}{4}(x-6) \quad\left[\Rightarrow y=-\frac{3}{4} x+\frac{29}{2}\right]\)
c.Coordinates of centre of circle \(Q\) are \(\left(0\right.\), their \(\left.\frac{29}{2}\right)\)
Equation of circle \(Q\) is \(x^2+\left(y \text {-their } \frac{29}{2}\right)^2=\left(\frac{5 \sqrt{5}}{2}\right)^2\left[=\frac{125}{4}\right]\)
\(x^2+(11-2)^2=100 \Rightarrow x^2=19\) and \(x^2+\left(11-\frac{29}{2}\right)^2=\frac{125}{4} \Rightarrow x^2=19\)
OR e.g. \(\frac{125}{4}-\left(y-\frac{29}{2}\right)^2+(y-2)^2=100 \Rightarrow 25 y=275 \Rightarrow y=11\)
d.$
x^2+\left(-\frac{3}{4} x+\frac{29}{2}-\frac{29}{2}\right)^2=\frac{125}{4}\left[\Rightarrow \frac{25}{16} x^2=\frac{125}{4} \Rightarrow x^2=20\right]
$
or \(y^2-29 y+199[=0]\)
\(x= \pm 2 \sqrt{5}\) or \(y=\frac{29 \mp 3 \sqrt{5}}{2}\)
\(y\left[=\left(-\frac{3}{4} \times \pm \sqrt{20}\right)+\frac{29}{2}\right]=\frac{29 \mp 3 \sqrt{5}}{2}\)