Exercise 5.6-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
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}
$