Exercise 5.5-Additional Problems - Chapter 5 Two Dimensional Analytical Geometry–II 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Problems
Question 1.
If a parabolic reflector is $20 \mathrm{~cm}$ in diameter and $5 \mathrm{~cm}$ deep, find the distance of the focus from the centre of the reflector.
Solution:

By the property of the parabolic reflector, the position of the bulb should be placed at the focus.
By taking the vertex at the origin the equation of the reflector is $y^2=4 a x$.
Let $P Q$ be the diameter of the reflector $P=(5,10)$
Since $P(5,10)$ lies on the parabola,
$
10^2=4 \mathrm{a} \times 5
$
ie., $100=20 \mathrm{a} \Rightarrow \mathrm{a}=5$
So the focus is at a distance of $5 \mathrm{~cm}$ from the vertex and focus is $(5,0)$.

Question 2.
The focus of a parabolic mirror is at a distance of $8 \mathrm{~cm}$ from its centre (vertex). If the mirror $25 \mathrm{~cm}$ deep, find the diameter of the mirror.
Solution:

Let the vertex be at the origin.
$
\mathrm{VF}=\mathrm{a}=8 \mathrm{~cm}
$
The equation of the parabola is
$
\mathrm{Y}^2=4 \mathrm{ax}=4(8) \mathrm{x}=32 \mathrm{x}
$
Depth of the mirror $=\mathrm{x}_1=25 \mathrm{~cm}$
So, radius is 0 .
$
\begin{aligned}
& \Rightarrow \mathrm{y}^2=32(25)=800 \\
& \mathrm{y}=\sqrt{800}=10 \sqrt{8}=10 \times 2 \sqrt{2}=20 \sqrt{2}=\text { Radius of the mirror } \\
& \therefore \text { Diameter of the mirror }=2 \times 20 \sqrt{2}=40 \sqrt{2} \mathrm{~cm} \text { of the mirror. }
\end{aligned}
$
Question 3.
A cable of a suspension bridge is in the form of a parabola whose span is $40 \mathrm{mts}$. The roadway is $5 \mathrm{mts}$ below the lowest point of the cable. If an extra support is provided across the cable $30 \mathrm{mts}$ above the ground level, find the length of the support if the height of the pillars are $55 \mathrm{mts}$.
Solution:

The lowest point on the cable is taken as the vertex and it is taken as the origin. Let $A B, C D$ be the pillars.
Span of parabola $=40 \mathrm{mts}=$ distance between $\mathrm{AB}$ and $C D$
$\mathrm{C}^{\prime} \mathrm{V}=\mathrm{VA}^{\prime}=20 \mathrm{mts}$
Height of each pillar $=55 \mathrm{mts} \Rightarrow \mathrm{AB}=55 \mathrm{mts}$
So, $\mathrm{A}^{\prime} \mathrm{B}=55-5=50 \mathrm{mts}$
Thus, the point $\mathrm{B}$ is $(20,50)$.
Equation of the parabola is $\mathrm{x}^2=4$ ay
Here, $B$ is a point on the parabola, $x^2=4$ ay
$(20)^2=4 \mathrm{a}(50) \Rightarrow 4 \mathrm{a}=\frac{20 \times 20}{50}=8$
$\therefore$ The equation is $\mathrm{x}^2=8 \mathrm{y}$
Let $P Q$ be the length of the extra support $R Q$.
$R Q=30, R^{\prime}=5 \Rightarrow R^{\prime} Q=25$
Let $V R^{\prime}$ be $\mathrm{x}_1 \therefore \mathrm{Q}$ is $\left(\mathrm{x}_1, 25\right)$.
$Q$ is a point on parabola
$
\begin{aligned}
& \mathrm{x}_1{ }^2=8 \times 25=200 \\
& \mathrm{x}_1=\sqrt{200}=10 \sqrt{2}
\end{aligned}
$
The entire length, $P Q=2 \mathrm{x}_1=20 \sqrt{2}$.mts.

Question 4.
A kho-kho player in a practice session while running realises that the sum of the distances from the kho-kho poles from him is always $8 \mathrm{~m}$. Find the equation of the path traced by him if the distance between the poles is $6 \mathrm{~m}$
Solution:

Give FP + F'P $=4$
ie., $2 a=8$ and
$\mathrm{FF}^{\prime}=2 \mathrm{ae}=6$
ie., $\mathrm{a}=4$ and $\mathrm{ae}=3$
$\mathrm{e}=\frac{a e}{a}=\frac{3}{4}$
$\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=16\left(1-\frac{9}{16}\right)=7$
So the equation of the path is an ellipse whose equation is $\frac{x^2}{16}+\frac{y^2}{7}=1$.