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

Updated On May 15, 2024

By SaraNextGen

Ex $5.5$
Question 1.
A bridge has a parabolic arch that is $10 \mathrm{~m}$ high in the centre and $30 \mathrm{~m}$ wide at the bottom. Find the height of the arch $6 \mathrm{~m}$ from the centre, on either sides.
Solution:
From the diagram, equation of the parabolic arch
$
x^2=-4 a y
$
Use the point $(15,-10)$ in (1)
$
\begin{aligned}
(1) \Rightarrow \quad 15^2 & =-4 a(-10) \\
225 & =40 a \\
\text { substitute } a & =\frac{225}{40} \text { in (1) } \\
(1) \Rightarrow \quad x^2 & =-4\left(\frac{225}{40}\right) y
\end{aligned}
$
Use the point $\left(6,-y_1\right)$ in (2)

$
\begin{aligned}
(2) \Rightarrow \quad(6)^2 & =-4\left(\frac{225}{40}\right)\left(-y_1\right) \\
\frac{36 \times 10}{225} & =y_1
\end{aligned}
$
$\therefore$ The required height $=10-\mathrm{y}_1=10-1.6=8.4 \mathrm{~m}$.
Question 2.
A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be $16 \mathrm{~m}$, and the height at the edge of the road must be sufficient for a truck $4 \mathrm{~m}$ high to clear if the highest point of the opening is to be $5 \mathrm{~m}$ approximately. How wide must the opening be?
Solution:
From the diagram,

$\mathrm{AA}^{\prime}=16 \mathrm{~m}, \mathrm{OA}=8 \mathrm{~m}, \mathrm{OB}=5 \mathrm{~m}$
$\therefore$ Equation of the ellipse is

$
\frac{x^2}{5^2}+\frac{y^2}{8^2}=1
$
$\therefore$ The point $\left(4, y_1\right)$ on the ellipse
$
\begin{aligned}
(1) \Rightarrow \frac{4^2}{5^2}+\frac{y_1^2}{8^2} & =1 \\
\frac{y_1^2}{64} & =1-\frac{16}{25} \\
y_1 & =\sqrt{\frac{9 \times 64}{25}}=\frac{3 \times 8}{5}=4.8 \mathrm{~m}
\end{aligned}
$
$\therefore$ The required wide for the opening is $2 \mathrm{y}_1=2(4.8)=9.6 \mathrm{~m}$

Question 3.
At a water fountain, water attains a maximum height of $4 \mathrm{~m}$ at horizontal distance of $0.5 \mathrm{~m}$ from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of $0.75 \mathrm{~m}$ from the point of origin.
Solution:
From the diagram
Equation of the path of water
$
x^2=-4 a y
$
Use the point $(0.5,-4)$ in (1)
$
\begin{aligned}
(0.5)^2 & =-4 a(-4) \\
\frac{(0.5)^2}{16}=a(\text { or }) \quad a & =\frac{0.25}{16} \text { substitute in (1) } \\
(1) \Rightarrow \quad x^2 & =-4\left[\frac{(0.25)}{16}\right] y
\end{aligned}
$
Use the point $\left(0.25,-y_1\right)$ in (2)
$
\begin{aligned}
(2) \Rightarrow(0.25)^2=-4\left[\frac{(0.25)}{16}\right]\left(-y_1\right) & \Rightarrow(0.25)^2=\frac{0.25}{4} y_1 \\
(0.25)^2 \times \frac{4}{0.25}=y_1 & \Rightarrow y_1=1 \mathrm{~m}
\end{aligned}
$
The refined height $=4-y_1=4-1=3 \mathrm{~m}$

Question 4.
An engineer designs a satellite dish with a parabolic cross section. The dish is $5 \mathrm{~m}$ wide at the opening, and the focus is placed $1.2 \mathrm{~m}$ from the vertex
(a) Position a coordinate system with the origin at the vertex and the $x$-axis on the parabola's axis of symmetry and find an equation of the parabola.
(b) Find the depth of the satellite dish at the vertex.
Solution:
From the diagram,

(a) Consider the satellite dish is open rightward parabola $\mathrm{y}^2=4 \mathrm{ax} \ldots \ldots \ldots \ldots$ (1)
Clearly a $=1.2 \mathrm{~m}$
(1) $\Rightarrow y^2=4(1.2)$ $y^2=4.8 x$
(b) Use the point $\left(\mathrm{x}_1, 2.5\right)$ in (1)
$(2.5)^2=4(1.2) \mathrm{x}_1$
$\frac{(2.5)^2}{4(1.2)}=\mathrm{y}_1$
$\mathrm{x}_1=1.3 \mathrm{~m}$
$\therefore$ The depth of the satellite dish at vertex is $1.3 \mathrm{~m}$
Question 5.
Parabolic cable of a $60 \mathrm{~m}$ portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every $6 \mathrm{~m}$ along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.
Solution:
From the diagram,

Equation of the suspension bridge
$
(x-h)^2=4 a(y-k)
$
But V $(0,3)$
$
x^2=4 a(y-3)
$
Use the point $(30,16)$ in (1)
$
\begin{aligned}
& 30^2=4 \mathrm{a}(16-3) \Rightarrow 900=13 \times 4 \mathrm{a} \\
& \frac{900}{13 \times 4}=\mathrm{a}
\end{aligned}
$

$\Rightarrow$
$
\begin{aligned}
& x^2=4\left(\frac{900}{13 \times 4}\right)(y-3) \\
& x^2=\left(\frac{900}{13}\right)(y-3)
\end{aligned}
$

(i) The length of the first vertical cable from the vertex is
Use $\left(6, y_1\right)$ in (2)
$
\begin{aligned}
& (2) \Rightarrow(6)^2=\frac{900}{13}\left(y_1-3\right) \\
& \frac{36 \times 13}{900}=y_1-3 \\
& 0.52=y_1-3 \\
& y_1=3.52 \mathrm{~m}
\end{aligned}
$
(ii) The length of the second vertical cable from the vertex is
Use the point $\left(12, y_2\right)$ in (2)
$
\begin{aligned}
& (2) \Rightarrow(12)^2=\frac{900}{13}\left(y_2-3\right) \\
& \frac{144 \times 13}{900}=y_2-3 \\
& 0.52=y_2-3 \\
& y_2=3.52 \mathrm{~m}
\end{aligned}
$
Question 6.
Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation $\frac{x^2}{30^2}-\frac{y^2}{44^2}=1$. The tower is $150 \mathrm{~m}$ tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
Solution:
From the diagram,equation of hyperbola is

Substitute $\left(x_1, 50\right)$ in (1)
$
\begin{aligned}
\frac{x_1^2}{30^2}-\frac{(50)^2}{44^2} & =1 \\
\frac{x_1^2}{30^2} & =1+\frac{(50)^2}{(44)^2} \\
\frac{x_1^2}{30^2} & =\frac{(44)^2+(50)^2}{(44)^2} \\
x_1^2 & =\frac{(30)^2\left[(44)^2+(50)^2\right]}{(44)^2} \\
x_1 & =\frac{30}{44} \sqrt{(44)^2+(50)^2} \\
x_1 & =45.41 \mathrm{~m} \\
\text { diameter } 2 x_1 & =2(45.41) \\
& =90.82 \mathrm{~m}
\end{aligned}
$
Sub $\left(x_2, 100\right)$ in (1)
$
\begin{aligned}
\frac{x_2^2}{(30)^2}-\frac{(100)^2}{44^2} & =1 \\
x_2 & =\frac{30}{44} \sqrt{(44)^2+(100)^2} \\
x_2 & =74.49 \mathrm{~m} \\
\text { diameter } & =2 x_1=2(74.49)=148.98 \mathrm{~m}
\end{aligned}
$

Question 7.
A rod of length $1.2 \mathrm{~m}$ moves with its ends always touching the coordinate axes. The locus of a point $\mathrm{P}$ on the rod, which is $0.3 \mathrm{~m}$ from the end in contact with $\mathrm{x}$-axis is an ellipse. Find the eccentricity.
Solution:
From the diagram,

(i) $\Delta^{\text {le }} \mathrm{OAB}$ be a right angle triangle.

(ii) $\angle \mathrm{APD}$ and $\angle \mathrm{PBC}$ are corresponding angles, so corresponding angles are equal.
$
\begin{array}{c|c}
\text { In } \Delta^{\text {le }} \mathbf{A D P} & \text { In } \Delta^{\text {le }} \mathbf{P C B} \\
\cos \theta=\frac{a d j}{h y p} & \sin \theta=\frac{o p p}{h y p} \\
\cos \theta=\frac{x_1}{0.9} & \sin \theta=\frac{y_1}{0.3}
\end{array}
$
We know that
$
\begin{array}{r}
\cos ^2 \theta+\sin ^2 \theta=1 \\
\frac{x_1^2}{(0.9)^2}+\frac{\left(y_1\right)^2}{(0.3)^2}=1
\end{array}
$
$\therefore$ The locus of $\mathrm{P}\left(x_1, y_1\right)$ is an ellipse.
$
\begin{aligned}
\frac{x^2}{(0.9)^2}+\frac{y^2}{(0.3)^2} & =1 \\
a^2 & =(0.9)^2 \\
b^2 & =(0.3)^2 \\
\text { Eccentricity } & =\sqrt{\frac{a^2-b^2}{a^2}} \\
& =\sqrt{\frac{(0.9)^2-(0.3)^2}{(0.9)^2}}=\sqrt{\frac{0.81-0.09}{0.81}}=\sqrt{\frac{0.72}{0.81}}=\sqrt{\frac{72}{81}}=\sqrt{\frac{8}{9}} \\
e & =\frac{2 \sqrt{2}}{3}
\end{aligned}
$

Question 8.
Assume that water issuing from the end of a horizontal pipe, $7.5 \mathrm{~m}$ above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position $2.5 \mathrm{~m}$ below the line of the pipe, the flow of water has curved outward $3 \mathrm{~m}$ beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground?
Solution:
From the diagram,

Equation of the water path is $\mathrm{x}^2=-4$ ay
Use the point $(3,-2.5)$ in (1)
$(3)^2=-4 \mathrm{a}(-2.5)$
$9=10 a$
$\mathrm{a}=\frac{9}{10}$ substituting in (1)
(1) $\Rightarrow x^2=-4 \frac{9}{10} y$
Use the point $\left(\mathrm{x}_1,-7.5\right)$ in (2)
$
\begin{aligned}
& (2) \Rightarrow \mathrm{x}_1^2=-4 \frac{9}{10}(-7.5) \Rightarrow \mathrm{x}_1^2=30\left(\frac{9}{10}\right) \\
& \mathrm{x}_1=\sqrt{3 \times 9} \\
& \mathrm{x}_1=3 \sqrt{3} \mathrm{~m}
\end{aligned}
$
$\therefore$ The water strikes the ground $3 \sqrt{3} \mathrm{~m}$ beyond the vertical line.
Question 9.
On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of $4 \mathrm{~m}$ when it is $6 \mathrm{~m}$ away from the point of projection. Finally it reaches the ground $12 \mathrm{~m}$ away from the starting point.
Find the angle of projection?
Solution:
From the diagram,

Equation of the parabolic path is
$
x^2=-4 a y
$
Use the point $(6,-4)$ in (1)
(1) $\Rightarrow(6)^2=16 \mathrm{a}$
$
\begin{aligned}
& \frac{36}{16}=10 \mathrm{a} \\
& \text { substitute } \mathrm{a}=\frac{9}{14} \text { in (1) } \\
& (1) \Rightarrow \mathrm{x}^2=-4\left(\frac{9}{4}\right) \mathrm{y} \\
& \mathrm{x}^2=-9 \mathrm{y} \ldots \ldots \ldots . \text { (2) }
\end{aligned}
$
Differentiate with respect to ' $\mathrm{x}$ '
$
2 x=-9 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-\frac{2 x}{9}
$
$\therefore$ The slope at $(-6,-4)$
$
\begin{aligned}
\tan \theta & =\frac{-2(-6)}{9} \Rightarrow \tan \theta=\frac{4}{3} \\
\theta & =\tan ^{-1}\left(\frac{4}{3}\right)
\end{aligned}
$

Question 10.
Points $\mathrm{A}$ and $\mathrm{B}$ are $10 \mathrm{~km}$ apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is $6 \mathrm{~km}$ closer to $\mathrm{A}$ than $\mathrm{B}$. Show that the location of the explosion is restricted to a particular curve and find an equation of it.
Solution:
From the diagram,

$\begin{aligned}
\mathrm{F}_1 \mathrm{P} \sim \mathrm{F}_2 \mathrm{P} & =6 \\
2 a & =6 \\
a & =3 \\
a e & =5 \Rightarrow e=\frac{5}{3} \\
b^2 & =a^2\left[e^2-1\right]=9\left[\frac{25}{9}-1\right] \\
& =9\left[\frac{25-9}{9}\right] \\
b^2 & =16
\end{aligned}$
$\therefore$ The required equation is
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1
$