Exercise 6.1 - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Two Dimensional Analytical Geometry
Ex 6.1
Question 1.
Find the locus of $\mathrm{P}$, if for all values of a, the co-ordinates of a moving point $\mathrm{P}$ is
(i) $(9 \cos \alpha, 9 \sin \alpha)$
(ii) $(9 \cos \alpha, 6 \sin \alpha)$
Solution:
(i) Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be the moving point.
We are given $\mathrm{h}=9 \cos \alpha$ and $\mathrm{k}=9 \sin \alpha$ and
$\Rightarrow \cos \alpha=\frac{h}{9}$ and $\sin \alpha=\frac{k}{9}$
We know $\cos ^2 \alpha+\sin ^2 \alpha=1 \Rightarrow\left(\frac{h}{9}\right)^2+\left(\frac{k}{9}\right)^2=1$
(i.e.) $\frac{h^2}{81}+\frac{k^2}{81}=1 \Rightarrow h^2+k^2=81$
$\therefore$ locus of the point is $x^2+y^2=81$
(ii) Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a moving point.
We are given $\mathrm{h}=9 \cos \alpha$ and $\mathrm{k}=6 \sin \alpha$
$
\begin{gathered}
\Rightarrow \cos \alpha=\frac{h}{9} \text { and } \sin \alpha=\frac{k}{6} \\
\cos ^2 \alpha+\sin ^2 \alpha=1 \Rightarrow\left(\frac{h}{9}\right)^2+\left(\frac{k}{6}\right)^2=1 \\
\therefore \frac{h^2}{81}+\frac{k^2}{36}=1
\end{gathered}
$
Locus of the point is $\frac{x^2}{81}+\frac{y^2}{36}=1$
Question 2.
Find the locus of a point $P$ that moves at a constant distant of
(i) Two units from the $x$-axis
(ii) Three units from the $y$-axis.
Solution:
(i) Let the point $(\mathrm{x}, \mathrm{y})$ be the moving point. Equation of a line at a distance of 2 units from $\mathrm{x}$-axis is $\mathrm{k}=2$ So the locus is $\mathrm{y}=2$

$
\begin{aligned}
& \text { (i.e.) } y-2=0 \\
& 435
\end{aligned}
$
(ii) Equation of a line at a distance of 3 units from $y$-axis is $\mathrm{h}=3$
So the locus is $x=3$ (i.e.) $x-3=0$
Question 3.
If $\theta$ is a parameter, find the equation of the locus of a moving point, whose coordinates are $\mathrm{x}-\mathrm{a} \cos ^3 \theta, \mathrm{y}$ $=\mathrm{a} \sin ^3 \theta$
Solution:
$
x=a \cos ^3 \theta \Rightarrow \cos ^3 \theta=\frac{x}{a} \Rightarrow \cos \theta=\left(\frac{x}{a}\right)^{1 / 3}
$
$
\begin{aligned}
& y=a \sin ^3 \theta \Rightarrow \sin ^3 \theta=\frac{y}{a} \\
& \Rightarrow \sin \theta=\left(\frac{y}{a}\right)^{1 / 3}
\end{aligned}
$
But $\cos ^2 \theta+\sin ^2 \theta=1 \Rightarrow\left\{\left(\frac{x}{a}\right)^{1 / 3}\right\}^2+\left\{\left(\frac{y}{a}\right)^{1 / 3}\right\}^2=1$
(i.e) $\frac{x^{2 / 3}}{a^{2 / 3}}+\frac{y^{2 / 3}}{a^{2 / 3}}=1 \Rightarrow x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$
Question 4.
Find the value of $k$ and $b$, if the points $P(-3,1)$ and $Q(2, b)$ lie on the locus of $x^2-5 x+k y=0$.
Solution:
Given $P(-3,1)$ lie on $x^2-5 x+k y=0$
$
\begin{aligned}
& \Rightarrow(-3)^2-5(-3)+\mathrm{k}(1)=0 \\
& 9+15+\mathrm{k}=0 \Rightarrow \mathrm{k}=-24
\end{aligned}
$
$\mathrm{Q}(2, \mathrm{~b})$ lie on $\mathrm{x}^2-5 \mathrm{x}+\mathrm{ky}=0$
$(2)^2-5(2)+\mathrm{k}(\mathrm{b})=0 \Rightarrow 4-5(2)-24 \mathrm{~b}=0$
$4-10-24 b=0 \Rightarrow-24 b=6 \Rightarrow b=6 /-24=-1 / 4$
Question 5.
A straight rod of length 8 units slides with its ends $\mathrm{A}$ and $\mathrm{B}$ always on the $\mathrm{x}$ and $\mathrm{y}$-axis respectively. Find the locus of the mid point of the line segment $\mathrm{AB}$.
Solution:

Let $P(h, k)$ be the moving point $A(a, 0)$ and $B(0, b) P$ is the mid point of $A B$
$
\Rightarrow \mathrm{P}=\left(\frac{a}{2}, \frac{b}{2}\right)
$

Given $\frac{a}{2}=h$ and $\frac{b}{2}=k \Rightarrow a=2 h$ and $b=2 k$
$\mathrm{OAB}$ is a right angled triangle $\Rightarrow \mathrm{OA}^2+\mathrm{OB}^2=\mathrm{AB}^2$
(i.e.) $a^2+b^2=8^2 \quad[\because \mathrm{AB}=8]$
$\Rightarrow(2 h)^2+(2 k)^2=8^2$ (i.e) $4 h^2+4 k^2=64$
$(\div$ by 4$) h^2+k^2=16$
So the locus of $P$ is $x^2+y^2=16$

Question 6.
Find the equation of the locus of a point such that the sum of the squares of the distance from the points $(3,5),(1,-1)$ is equal to 20 .
Solution:
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be the moving point
Let the given point be $A(3,5)$ and $B(1,-1)$
We are given $\mathrm{PA}^2+\mathrm{PB}^2=20$
$
\begin{aligned}
& \Rightarrow(\mathrm{h}-3)^2+(\mathrm{k}-5)^2+(\mathrm{h}-1)^2+(\mathrm{k}+1)^2=20 \\
& \Rightarrow \mathrm{h}^2-6 \mathrm{~h}+9+\mathrm{k}^2-10 \mathrm{k}+25+\mathrm{h}^2-2 \mathrm{~h}+1+\mathrm{k}^2+2 \mathrm{k}+1=20 \\
& \text { (i.e.) } 2 \mathrm{~h}^2+2 \mathrm{k}^2-8 \mathrm{~h}-8 \mathrm{k}+36-20=0 \\
& 2 \mathrm{~h}^2+2 \mathrm{k}^2-8 \mathrm{~h}-8 \mathrm{k}+16=0 \\
& (\div \mathrm{by} 2) \mathrm{h}^2+\mathrm{k}^2-4 \mathrm{~h}-4 \mathrm{k}+8=0
\end{aligned}
$
So the locus of $P$ is $x^2+y^2-4 x-4 y+8=0$
Question 7.
Find the equation of the locus of the point $P$ such that the line segment $A B$, joining the points $A(1,-6)$ and $B(4,-2)$, subtends a right angle at $P$.
Solution:
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be the moving point

Given $A(1,-6)$ and $B(4,-2)$,
Since $\triangle \mathrm{APB}=90^{\circ}, \mathrm{PA}^2+\mathrm{PB}^2=\mathrm{AB}^2$
(i.e.) $(\mathrm{h}-1)^2+(\mathrm{k}+6)^2+(\mathrm{h}-4)^2+(\mathrm{k}+2)^2=(4-1)^2+(-2+6)^2$
(i.e) $\mathrm{h}^2+1-2 \mathrm{~h}+\mathrm{k}^2+36+12 \mathrm{k}+\mathrm{h}^2+16-8 \mathrm{~h}+\mathrm{k}^2+4+4 \mathrm{k}=3^2+4^2=25$
$2 \mathrm{~h}^2+2 \mathrm{k}^2-10 \mathrm{~h}+16 \mathrm{k}+57-25=0$
$2 \mathrm{~h}^2+2 \mathrm{k}^2-10 \mathrm{~h}+16 \mathrm{k}+32=0$
$(\div \mathrm{by} 2) \mathrm{h}^2+\mathrm{k}^2-5 \mathrm{~h}+8 \mathrm{k}+16=0$
So the locus of $P$ is $x^2+y^2-5 x+8 y+16=0$
Question 8.
If $O$ is origin and $R$ is a variable point on $y^2=4 x$, then find the equation of the locus of the mid-point of the line segment OR.
Solution:
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be the moving point
We are given $O(0,0)$. Let $R=(a, b)$
Mid point of $\mathrm{OR}=\left(\frac{a}{2}, \frac{b}{2}\right) \Rightarrow\left(\frac{a}{2}, \frac{b}{2}\right)=(h, k)$ $\Rightarrow a=2 h, b=2 k$
Substituting a, b values is $y^2=4 x$
we get $(2 \mathrm{k})^2=4(2 \mathrm{~h})$
(i.e) $4 \mathrm{k}^2=8 \mathrm{~h}$
$(\div$ by 4$) \mathrm{k}^2=2 \mathrm{~h}$
So the locus of $P$ is $y^2=2 x$
Question 9.
The coordinates of a moving point $\mathrm{P}$ are $\left(\frac{a}{2}(\operatorname{cosec} \theta+\sin \theta), \frac{b}{2}(\operatorname{cosec} \theta-\sin \theta)\right)$ where I is a variable parameter. Show that the equation of the locus $\mathbf{P}$ is $b^2 x^2-a^2 y^2=a^2 b^2$.

Solution:
Let $\mathrm{P}(h, k)$ be the moving point
We are given $\mathrm{P}=\left[\frac{a}{2}(\operatorname{cosec} \theta+\sin \theta), \frac{b}{2}(\operatorname{cosec} \theta-\sin \theta)\right]$
$
\begin{aligned}
& \Rightarrow h=\frac{a}{2}(\operatorname{cosec} \theta+\sin \theta)(\text { i.e. }) a[\operatorname{cosec} \theta+\sin \theta] \\
& \Rightarrow \operatorname{cosec} \theta+\sin \theta=\frac{2 h}{a} \\
& \frac{b}{2}(\operatorname{cosec} \theta-\sin \theta)=k \Rightarrow \operatorname{cosec} \theta-\sin \theta=\frac{2 k}{b}
\end{aligned}
$
(1) $+(2) \Rightarrow 2 \operatorname{cosec} \theta=\frac{2 h}{a}+\frac{2 k}{b}$
$(\div$ by 2$) \Rightarrow \operatorname{cosec} \theta=\frac{h}{a}+\frac{k}{b}$
(1) $-(2) \Rightarrow 2 \sin \theta=\frac{2 h}{a}-\frac{2 k}{b}(\div b y 2) \Rightarrow \sin \theta=\frac{h}{a}-\frac{k}{b}$.
But $\operatorname{cosec} \theta \sin \theta=1$, So from (3) $\&(4) \Rightarrow\left(\frac{h}{a}+\frac{k}{b}\right)\left(\frac{h}{a}-\frac{k}{b}\right)=1$ $\Rightarrow \frac{h^2}{a^2}-\frac{k^2}{b^2}=1$
So the locus of $(h, k)$ is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
(i.e) $\frac{b^2 x^2-a^2 y^2}{a^2 b^2}=1 \Rightarrow b^2 x^2-a^2 y^2=a^2 b^2$

Question 10.
If $P(2,-7)$ is a given point and $Q$ is a point on $2 x^2+9 y^2=18$, then find the equations of the locus of the mid-point of $P Q$.
Solution:
$\mathrm{P}=(2,-7)$; Let $(\mathrm{h}, \mathrm{k})$ be the moving point $\mathrm{Q}=(\mathrm{a}, \mathrm{b})$
Mid point of $P Q=\left(\frac{2+a}{2}, \frac{-7+b}{2}\right)=(h, k)$ say
$
\begin{aligned}
& \frac{a+2}{2}=h, \quad \frac{b-7}{2}=k \\
& \Rightarrow \mathrm{a}=2 \mathrm{~h}-2, \\
& \mathrm{~b}=2 \mathrm{k}+1
\end{aligned}
$
$Q$ is a point on $2 x^2+9 y^2=18$ (i.e) $(a, b)$ is on $2 x^2+9 y^2=18$
$\Rightarrow 2(2 \mathrm{~h}-2)^2+9(2 \mathrm{k}+7)^2=18$
(i.e) $2\left[4 \mathrm{~h}^2+4-8 \mathrm{~h}\right]+9\left[4 \mathrm{k}^2+49+28 \mathrm{k}\right]-18=0$
(i.e) $8 \mathrm{~h}^2+8-16 \mathrm{~h}+36 \mathrm{k}^2+441+252 \mathrm{k}-18=0$
$
8 \mathrm{~h}^2+36 \mathrm{k}^2-16 \mathrm{~h}+252 \mathrm{k}+431=0
$
The locus is $8 x^2+36 y^2-16 x+252 y+431=0$

Question 11.
If $R$ is any point on the $x$-axis and $Q$ is any point on the $y$-axis and Pis a variable point on $R Q$ with $R P=$ $b, P Q=a$. then find the equation of locus of $P$.
Solution:
$P=(x, 0), Q=(0, y), R(h, k)$ be a point on $R Q$ such that $P R: R Q=b: a$
$
\begin{aligned}
& \therefore \mathrm{P}=\left(\frac{a x}{a+b}, \frac{b y}{a+b}\right) \Rightarrow\left(\frac{a x}{a+b}, \frac{b y}{a+b}\right)=(h, k) \\
& \Rightarrow a x=(a+b) h \Rightarrow x=\frac{a+b}{a} h \\
& b y=(a+b) k \Rightarrow y=\frac{a+b}{b} k
\end{aligned}
$
From the right angled triangle $\mathrm{OQR}, \mathrm{OR}^2+\mathrm{OQ}^2=\mathrm{QR}^2$

$
\begin{aligned}
& \text { (i.e) } \mathrm{X}^2+\mathrm{Y}^2=(a+b)^2 \\
& \Rightarrow\left[\frac{a+b}{a} h\right]^2+\left[\frac{a+b}{b} k\right]^2=(a+b)^2 \\
& \Rightarrow \frac{(a+b)^2}{a^2} h^2+\frac{(a+b)^2}{b^2} k^2=(a+b)^2 \\
& \div b y(a+b)^2, \frac{h^2}{a^2}+\frac{k^2}{b^2}=1 \\
& \text { Locus is } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{aligned}
$
Question 12.
If the points $P(6,2)$ and $Q(-2,1)$ and $R$ are the vertices of a $\triangle P Q R$ and $R$ is the point on the locus $y=x^2$ $-3 x+4$, then find the equation of the locus of centroid of $\triangle P Q R$.

Solution:
$P(6,2), Q(-2,1)$. Let $R=(a, b)$ be a point on $y=x^2-3 x+4$.
Centroid of $\triangle \mathrm{PQR}$ is $\left(\frac{6-2+a}{3}, \frac{2+1+b}{3}\right)$
$
\begin{aligned}
& \text { (i.e) }\left(\frac{4+a}{3}, \frac{3+b}{3}\right)=(h, k) \\
& \frac{4+a}{3}=h \Rightarrow a=3 h-4 \\
& \frac{3+b}{3}=k \Rightarrow b=3 k-3
\end{aligned}
$
But (a, b) is a point on $y=x^2-3 x+4$
$
\mathrm{b}=\mathrm{a}^2-3 \mathrm{a}+4
$
(i.e) $3 \mathrm{k}-3=(3 \mathrm{~h}-4)^2-3(3 \mathrm{~h}-4)+4$
(i.e) $3 \mathrm{k}-3=9 \mathrm{~h}^2+16-24 \mathrm{~h}-9 \mathrm{~h}+12+4$
$
\Rightarrow 9 \mathrm{~h}^2-24 \mathrm{~h}-9 \mathrm{~h}+32-3 \mathrm{k}+3=0
$
(i.e) $9 \mathrm{~h}^2-33 \mathrm{~h}-3 \mathrm{k}+35=0$,
Locus of $(h, k)$ is $9 x^2-33 x-3 y+35=0$
Question 13.
If $Q$ is a point on the locus of $x^2+y^2+4 x-3 y+7=0$ then find the equation of locus of $P$ which divides segment $O Q$ externally in the ratio $3: 4$, where $O$ is origin.
Solution:
Let $(\mathrm{h}, \mathrm{k})$ be the moving point $\mathrm{O}=(0,0)$;
Let $P Q=(a, b)$ on $x^2+y^2+4 x-3 y+7=0$
$\mathrm{P}$ divides $\mathrm{OQ}$ externally in the ratio $3: 4 \Rightarrow \mathrm{P}=\left(\frac{3(a)-0}{-1}, \frac{3(b)-0}{-1}\right)=(-3 a,-3 b)=(h, k)$ $a=-\frac{h}{3}$ and $b=-\frac{k}{3}$
But $(a, b)$ is on $x^2+y^2+4 x-3 y+7=0$
$
\Rightarrow\left(-\frac{h}{3}\right)^2+\left(-\frac{k}{3}\right)^2+4\left(-\frac{h}{3}\right)-3\left(-\frac{k}{3}\right)+7=0
$
(i.e) $\frac{h^2}{9}+\frac{k^2}{9}-\frac{4 h}{3}+k+7=0$
$h^2+k^2-12 h+9 k+63=0$, Locus of $(h, k)$ is $x^2+y^2-12 x+9 y+63=0$

Question 14.
Find the points on the locus of points that are 3 units from $\mathrm{x}$ - axis and 5 units from the point $(5,1)$.

Solution:
A line parallel to $x$-axis is of the form $\mathrm{y}=\mathrm{k}$. Here $\mathrm{k}=3 \Rightarrow \mathrm{y}=3$
A point on this line is taken as $P(a, 3)$. The distance of $P(a, 3)$ from $(5,1)$ is given as 5 units
$
\begin{aligned}
& \Rightarrow(a-5)^2+(3-1)^2=5^2 \\
& a^2+25-10 a+9+1-6=25 \\
& a^2-10 a+25+4-25=0 \\
& a^2-10 a+4=0 \\
& \quad a=\frac{10 \pm \sqrt{100-4(1)(4)}}{2(1)}=\frac{10 \pm \sqrt{84}}{2}=\frac{10 \pm 2 \sqrt{21}}{2}=5 \pm \sqrt{21}
\end{aligned}
$
So the points are $(5+\sqrt{21}, 3),(5-\sqrt{21}, 3)$.
Question 15
The sum of the distance of a moving point from the points $(4,0)$ and $(-4,0)$ is always 10 units. Find the equation of the locus of the moving point.
Solution:
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a moving point
Here $A=(4,0)$ and $\mathrm{B}=(-4,0)$
Given $\mathrm{PA}+\mathrm{PB}=10$

$
\begin{aligned}
& \Rightarrow \sqrt{(h-4)^2+k^2}+\sqrt{(h+4)^2+k^2}=10 \\
& \Rightarrow \sqrt{(h-4)^2+k^2}=10-\sqrt{(h+4)^2+k^2}
\end{aligned}
$
Squaring on both sides $(h-4)^2+k^2=100+(h+4)^2+k^2-20 \sqrt{(h+4)^2+k^2}$
(i.e.) $h^2+16-8 h+k^2=100+h^2+16+8 h+k^2-20 \sqrt{(h+4)^2+k^2}$
$
\begin{aligned}
& \Rightarrow-16 h-100=-20 \sqrt{(h+4)^2+k^2} \\
& (\div \text { by }-4) 4 h+25=5 \sqrt{(h+4)^2+k^2}
\end{aligned}
$
Again Squaring on both sides we get,
$
\begin{aligned}
& (4 h+25)^2=25\left[(h+4)^2+k^2\right] \\
& \text { (i.e) } 16 h^2+625+200 h=25\left[h^2+8 h+16+k^2\right] \\
& \Rightarrow 16 h^2+625+200 h-25 h^2-200 h-400-25 k^2=0 \\
& -9 h^2-25 k^2+225=0 \\
& \Rightarrow 9 h^2+25 k^2=225 \\
& \frac{9 h^2}{225}+\frac{25 k^2}{225}=1 \\
& \text { (i.e) } h^2 / 25+k^2 / 9=1
\end{aligned}
$
So the locus is $\frac{x^2}{25}+\frac{y^2}{9}=1$