Exercise 5.6-Additional Problems - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 5.6-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
Choose the most appropriate answer:
Question 1.
The parabola $\mathrm{y}^2=4 \mathrm{ax}$ passes through the point $(2,-6)$, the the length of its latus rectum is
(a) 9
(b) 16
(c) 18
(d) 6
Solution:
(c) 18
Hint:
The given parabola is $y^2=4 \mathrm{ax}$
Since it passes through $(2,-6)$
$
\begin{aligned}
& \therefore(-6)^2=4 \mathrm{a}(2) \\
& \Rightarrow 36=8 \mathrm{a} a=\frac{36}{8}=\frac{9}{2}
\end{aligned}
$
Length of the latus rectum $=4 \mathrm{a}=4\left(\frac{9}{2}\right)=18$
Question 2.
The vertex of the parabola $x^2+12 x-9 y=0$ is
(a) $(6,-1)$
(b) $(-6,4)$
(c) $(6,4)$
(d) $(-6,-4)$
Solution:
(d) $(-6,-4)$
Hint:
Given parabola is $x^2+12 x-9 y=0$
$
\begin{aligned}
& \Rightarrow\left(\mathrm{x}^2+12 \mathrm{x}+36\right)-36-9 \mathrm{y}=0 \\
& \Rightarrow(\mathrm{x}+6)^2=9 \mathrm{y}+36 \Rightarrow(\mathrm{x}+6)^2=9(\mathrm{y}+4) \\
& \therefore \text { Vertex }=(-6,-4)
\end{aligned}
$
Question 3.
The length of the major axis of an ellipse is three times the length of minor axis, its eccentricity is
(a) $\frac{1}{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) $\frac{1}{\sqrt{2}}$
(d) $\frac{2 \sqrt{2}}{3}$
Solution:
(d) $\frac{2 \sqrt{2}}{3}$

Hint:
Length of major axis $=2 \mathrm{a}$ and Length of minor axis $=2 \mathrm{~b}$
$
\begin{aligned}
& 2 \mathrm{a}=3(2 \mathrm{~b}) \Rightarrow \mathrm{a}=3 \mathrm{~b} \\
& \text { Now } b^2=a^2\left(1-e^2\right) \Rightarrow e^2=\frac{a^2-b^2}{9 b^2} \\
& \Rightarrow e^2=\frac{9 b^2-b^2}{9 b^2}=\frac{8}{9} ; e=\frac{2 \sqrt{2}}{3}
\end{aligned}
$
Question 4.
The eccentricity of the ellipse $9 x^2+4 y^2=36$ is
(a) $\sqrt{\frac{5}{3}}$
(b) $\sqrt{\frac{3}{5}}$
(c) $\frac{\sqrt{3}}{5}$
(d) $\frac{\sqrt{5}}{3}$
Solution:
(d) $\frac{\sqrt{ } 5}{3}$
Hint:
Given equation of the ellipse $9 x^2+4 y^2=36$
On dividing both the sides by 36 , we get
$
\begin{aligned}
& \quad \frac{x^2}{4}+\frac{y^2}{9}=1 . \text { Here } a<b \\
& \therefore \text { Eccentricity is given by } \\
& \qquad a^2=b^2\left(1-e^2\right) \Rightarrow \frac{a^2}{b^2}=1-e^2 \\
& \Rightarrow \quad e^2=1-\frac{a^2}{b^2}=\frac{b^2-a^2}{b^2}=\frac{9-4}{9}=\frac{5}{9}
\end{aligned}
$

$
e=\frac{\sqrt{5}}{3}
$
Question 5.
$\mathrm{S}$ and $\mathrm{T}$ are the foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\mathrm{B}$ is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\frac{\sqrt{3}}{2}$
Solution:
(c) $\frac{1}{2}$

Hint:
$\mathrm{S}$ and $\mathrm{T}$ are the foci of the ellipse
$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\
& \therefore \quad \mathrm{S}=(a e, 0) \\
& \text { and } \quad \mathrm{T}=(-a e, 0) \text { and } \\
& \mathrm{B}=(0, b) \\
& \because \mathrm{STB} \text { is an equilateral triangle } \\
& \therefore \quad \mathrm{ST}^2=\mathrm{TB}^2 \\
& \Rightarrow(a e+a e)^2+(0-0)^2=(-a e-0)^2+(0-b)^2 \\
& \Rightarrow \quad(2 a e)^2=a^2 e^2+b^2 \\
& \Rightarrow \quad 4 a^2 e^2=a^2 e^2+b^2 \Rightarrow 3 a^2 e^2=b^2 \\
& \Rightarrow \quad 3 a^2 e^2=a^2\left(1-e^2\right) \quad\left[b^2=a^2\left(1-e^2\right)\right] \\
& \Rightarrow \quad 4 a^2 e^2=a^2 \quad \Rightarrow \quad e^2=\frac{1}{4} \quad \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
$

Question 6.
The sum of the focal distances from any point of the ellipse $9 x^2+16 y^2=144$ is
(a) 32
(b) 18
(c) 16
(d) 8
Solution:
(d) 8
Hint:
We have, $9 x^2+16 y^2=144$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$ (Dividing both sides by 144 )
Here $a^2=16, b^2=9 \Rightarrow a=4, b=3$
Length of major axis $=2 a=2(4)=8$
Since the sum of the focal distances from any point on the ellipse is equal to its major axis
$\therefore$ Required sum $=8$
Question 7.
If the eccentricities of two ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are equal then $\frac{a}{b}=$
(a) $\frac{5}{13}$
(b) $\frac{6}{13}$
(c) $\frac{13}{5}$
(d) $\frac{13}{6}$
Solution:
(c) $\frac{13}{5}$

Hint:
Let $e_1$ and $e_2$ be the eccentricities of the ellipses $\frac{x^2}{169}+\frac{y^2}{25}=1$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ respectively
$\because$ The eccentricities of the two ellipses are equal,
$
\begin{aligned}
& \therefore \quad e_1=e_2 \Rightarrow e_1^2=e_2^2 \\
& \Rightarrow \frac{a_1^2-b_1^2}{a_1^2}=\frac{a_2^2-b_2^2}{a_2^2} \quad\left[\because b^2=a^2\left(1-e^2\right) \Rightarrow e^2=\frac{a^2-b^2}{a^2}\right] \\
& \Rightarrow \frac{169-25}{169}=\frac{a^2-b^2}{a^2} \Rightarrow \frac{144}{169}=\frac{a^2-b^2}{a^2} \\
& \Rightarrow \quad 144 a^2=169 a^2-169 b^2 \Rightarrow 25 a^2=169 b^2 \\
& \Rightarrow \quad \frac{a^2}{b^2}=\frac{169}{25} \Rightarrow\left(\frac{a}{b}\right)^2=\frac{169}{25} \Rightarrow \frac{a}{b}=\sqrt{\frac{169}{25}}=\frac{13}{5}
\end{aligned}
$
Question 8.
Equation of the hyperbola, whose eccentricity $\frac{3}{2}$ and foci at $(\pm 2,0)$ is
(a) $\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
(b) $\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}$
(c) $\frac{x^2}{4}-\frac{y^2}{9}=1$
(d) None of these
Solution:
(a) $\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
Hint:
Eccentricity e $=\frac{c}{a}=\frac{3}{2}$
Foci are $(\pm c, 0)=(\pm 2,0)($ Given $c=2)$
$
\begin{aligned}
\therefore \frac{c}{a} & =\frac{3}{2} \Rightarrow \frac{2}{a}=\frac{3}{2} \Rightarrow a=\frac{4}{3} \\
\text { Again } c^2 & =a^2+b^2 \Rightarrow b^2=c^2-a^2 \\
\Rightarrow b^2 & =(2)^2-\left(\frac{4}{3}\right)^2=4-\frac{16}{9}=\frac{20}{9}
\end{aligned}
$
Now, $a^2=\frac{16}{9}$ and $b^2=\frac{20}{9}$

$\begin{aligned}
& \therefore \text { Equation of hyperbola is } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\
& \Rightarrow \frac{x^2}{\frac{16}{9}}-\frac{y^2}{\frac{20}{9}} \Rightarrow \frac{9 x^2}{16}-\frac{9 y^2}{20}=1 \Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}
\end{aligned}$

Question 9.
If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and if $e_2$ is the eccentricity of the hyperbola $9 x^2-16 y^2$ $=144$, then $e_1 e_2$ is
(a) $\frac{16}{25}$
(b) 1
(c) Greater than 1
(d) Less than $\frac{1}{2}$
Solution:
(b) 1
Hint:
Equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$
Here, $\quad a^2=25, \quad b^2=9$
Now, $\quad b^2=a^2\left(1-e_1^2\right) \Rightarrow 9=25\left(1-e_1^2\right)$
$\Rightarrow \quad e_1^2=\frac{16}{25} ; e_1=\frac{4}{5}$
Equation of hyperbola is $\frac{x^2}{25}+\frac{y^2}{9}=1$
Here,
$
a^2=16, \quad b^2=9
$
$
b^2=a^2\left(e_2^2-1\right) \Rightarrow 9=16\left(e_2^2-1\right)
$
$
\begin{aligned}
\Rightarrow & e_2^2-1 & =\frac{9}{16} \Rightarrow e_2^2=\frac{25}{16} ; e_2=\frac{5}{4} \\
\therefore & e_1 e_2 & =\frac{4}{5} \times \frac{5}{4}=1
\end{aligned}
$

Question 10.
The point of intersection of the tangents at $t_1=t$ and $t_2=3 \mathrm{t}$ to the parabola $y^2=8 x$ is
(a) $\left(6 \mathrm{t}^2, 8 \mathrm{t}\right)$
(b) $\left(8 t, 6 t^2\right)$
(c) $\left(\mathrm{t}^2, 4 \mathrm{t}\right)$
(d) $\left(4 \mathrm{t}, \mathrm{t}^2\right)$
Solution:
(a) $\left(6 t^2, 8 t\right)$
Hint:
Point of intersection of the tangents at $t_1$ and $t_2$ to $y^2=4 a x$ is $\left[a t_1 t_2, a\left(t_1+t_2\right)\right]$
Here, $t_1=t, t_2=3 t, a=$ latex $] \backslash \operatorname{frac}\{8\}\{4\}[/$ latex $]=2$
So a $t_1 t_2=2(t)(3 t)=6 t^2$
$
\begin{aligned}
& a\left(t_1+t_2\right)=2(t+3 t)=8 t \\
& \therefore \text { Point }=\left(6 t^2, 8 t^2\right)
\end{aligned}
$