Exercise 5.5-Additional Problems - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 5.5-Additional Problems - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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 $\frac{x^2}{16}+\frac{y^2}{7}=1$.