Conic Sections
Miscellaneous Exercise Solutions
Q1
1. Parabolic reflector: 20 cm diameter, 5 cm deep. Find the focus.
Setup
Let vertex be at origin $(0,0)$. Axis along x-axis. Eq: $y^2 = 4ax$.
Coordinates
Diameter is 20, Depth is 5. The edge point is $(5, 10)$ or $(5, -10)$.
Substitute
Put $(5, 10)$ in eq: $(10)^2 = 4a(5) \Rightarrow 100 = 20a$.
Solve a
$a = 5$. Focus is at $(a, 0)$.
Focus is at (5, 0), i.e., 5cm from vertex.
Q2
2. Parabolic arch: 10m high, 5m wide at base. How wide is it 2m from the vertex?
Setup
Vertex at origin $(0,0)$, axis vertical downwards. Eq: $x^2 = -4ay$.
Point
Base width 5m, height 10m. End point is $(2.5, -10)$.
Find a
$(2.5)^2 = -4a(-10) \Rightarrow 6.25 = 40a \Rightarrow 4a = \frac{6.25}{10} = 0.625$. Eq: $x^2 = -0.625y$.
Target
Width at 2m from vertex (i.e., $y = -2$).
Calculate
$x^2 = -0.625(-2) = 1.25 \Rightarrow x = \sqrt{1.25} \approx 1.118$.
Width = $2x \approx 2.23$ m
Q3
3. Suspension bridge cable: Span 100m. Longest wire 30m, shortest 6m. Find wire length 18m from middle.
Coordinate
Vertex at lowest point of cable. Let vertex be $(0, 6)$ to match physical height, or shift origin. Let’s shift origin to lowest point $(0,0)$.
Logic
Equation relative to vertex: $x^2 = 4ay$.
Endpoint
Span 100m $\Rightarrow x=50$. Rise = $30 – 6 = 24$m. Point $(50, 24)$.
Find 4a
$50^2 = 4a(24) \Rightarrow 2500 = 96a \Rightarrow 4a = \frac{2500}{24} = \frac{625}{6}$.
Target
At $x=18$, find $y$. $18^2 = \frac{625}{6}y \Rightarrow 324 \times 6 = 625y \Rightarrow y = 3.11$.
Total Len
Add the base height (6m). $3.11 + 6 = 9.11$.
Length = 9.11 m
Q4
4. Semi-ellipse arch: 8m wide, 2m high. Find height 1.5m from one end.
Eq
Width $2a=8 \Rightarrow a=4$. Height $b=2$. $\frac{x^2}{16} + \frac{y^2}{4} = 1$.
Position
1.5m from end ($x=4$) means $x = 4 – 1.5 = 2.5$.
Substitute
$\frac{2.5^2}{16} + \frac{y^2}{4} = 1 \Rightarrow \frac{6.25}{16} + \frac{y^2}{4} = 1$.
Solve
$\frac{y^2}{4} = 1 – 0.3906 = 0.6094 \Rightarrow y^2 = 2.4375 \Rightarrow y \approx 1.56$.
Height $\approx$ 1.56 m
Q5
5. Rod length 12cm moving on axes. Locus of point P, 3cm from x-axis end.
Setup
Ends $A(a,0)$ and $B(0,b)$. Length $AB=12$. Point $P(x,y)$ divides AB.
Ratio
$AP = 3$. Since Total=12, $PB = 9$. Ratio $AP:PB = 1:3$.
Section
$x = \frac{1(0) + 3(a)}{4} \Rightarrow a = \frac{4x}{3}$.
$y = \frac{1(b) + 3(0)}{4} \Rightarrow b = 4y$.
$y = \frac{1(b) + 3(0)}{4} \Rightarrow b = 4y$.
Condition
$a^2 + b^2 = 12^2 = 144$. Substitute a and b.
Locus
$(\frac{4x}{3})^2 + (4y)^2 = 144 \Rightarrow \frac{16x^2}{9} + 16y^2 = 144$.
Simplify
Divide by 16: $\frac{x^2}{9} + y^2 = 9$ or $\frac{x^2}{81} + \frac{y^2}{9} = 1$.
$\frac{x^2}{81} + \frac{y^2}{9} = 1$ (Ellipse)
Q6
6. Area of triangle formed by vertex of $x^2 = 12y$ and ends of latus rectum.
Parabola
$x^2 = 4ay \Rightarrow 4a = 12 \Rightarrow a = 3$. Focus $(0, 3)$.
Latus Rectum
Passes through focus $y=3$. Endpoints: Substitute $y=3 \Rightarrow x^2 = 36 \Rightarrow x = \pm 6$. Points $(\pm 6, 3)$.
Vertices
Triangle vertices: $(0,0), (6,3), (-6,3)$.
Area
Base width = $6 – (-6) = 12$. Height = 3.
Area $= \frac{1}{2} \times 12 \times 3 = 18$.
Area $= \frac{1}{2} \times 12 \times 3 = 18$.
18 sq. units
Q7
7. Man running: Sum of distances from two flag posts is 10m. Distance between posts 8m. Find equation.
Definition
Locus of point where sum of distances to two fixed points (foci) is constant is an Ellipse.
Values
$2a = 10 \Rightarrow a = 5$. Distance between foci $2c = 8 \Rightarrow c = 4$.
Find b
$b^2 = a^2 – c^2 = 25 – 16 = 9$.
Equation
Assuming center at origin, foci on x-axis: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
$\frac{x^2}{25} + \frac{y^2}{9} = 1$
Q8
8. Equilateral triangle inscribed in $y^2 = 4ax$, one vertex at vertex of parabola. Find side length.
Symmetry
Due to symmetry, angle of side with x-axis is $30^\circ$.
Line Eq
Line passing through origin with angle $30^\circ$: $y = x \tan 30^\circ = \frac{x}{\sqrt{3}}$.
Intersection
Substitute line into parabola: $(\frac{x}{\sqrt{3}})^2 = 4ax \Rightarrow \frac{x^2}{3} = 4ax$.
Solve
$x = 12a$. Then $y = \frac{12a}{\sqrt{3}} = 4\sqrt{3}a$.
Side Length
Distance from origin $(0,0)$ to $(12a, 4\sqrt{3}a)$.
$L = \sqrt{(12a)^2 + (4\sqrt{3}a)^2} = \sqrt{144a^2 + 48a^2} = \sqrt{192a^2}$.
$L = \sqrt{(12a)^2 + (4\sqrt{3}a)^2} = \sqrt{144a^2 + 48a^2} = \sqrt{192a^2}$.
$8a\sqrt{3}$