Exercise 6.1-Additional Questions - Chapter 6 - Two Dimensional Analytical Geometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
If the sum of the distance of a moving point in a plane from the axis is 1 , then find the locus of the point. Solution:
Let coordinates of a moving point $\mathrm{P}$ be $(\mathrm{x}, \mathrm{y})$.
Given that the sum of the distances from the axis to the point is always 1 .

$
\begin{aligned}
& \therefore|\mathrm{x}|+|\mathrm{y}|=1 \Rightarrow \mathrm{x}+\mathrm{y}=1 \\
& \Rightarrow-\mathrm{x}-\mathrm{y}=1 \Rightarrow \mathrm{x}-\mathrm{y}=1
\end{aligned}
$
Hence, these equations give us the locus of the point $\mathrm{P}$ which is a square.
Question 2.
A point moves so that square of its distance from the point $(3,-2)$ is numerically equal to its distance from the line $5 x-12 y=3$. The equation of its locus is
Solution:
The given equation of line is $5 x-12 y=3$ and the given point is $(3,-2)$.
Let $(a, b)$ be any moving point.
$\therefore$ Distance between $(a, b)$ and the point $(3,-2)=\sqrt{(a-3)^2+(b+2)^2}$ and the distance 0 $(a, b)$ from the line $5 x-12 y=3$ is $\left|\frac{5 a-12 b-3}{\sqrt{25+144}}\right|=\left|\frac{5 a-12 b-3}{13}\right|$
According to the question, we have $\left[\sqrt{(a-3)^2+(b+2)^2}\right]^2=\left|\frac{5 a-12 b-3}{13}\right|$
$
\Rightarrow(a-3)^2+(b+2)^2=\frac{5 a-12 b-3}{13}
$
Taking numerical values only, we have $(a-3)^2+(b+2)^2=\frac{5 a-12 b-3}{13}$
$
\begin{aligned}
& \Rightarrow a^2-6 a+9+b^2+4 b+4=\frac{5 a-12 b-3}{13} \\
& \Rightarrow a^2+b^2-6 a+4 b+13=\frac{5 a-12 b-3}{13} \\
& \Rightarrow 13 a^2+13 b^2-78 a+52 b+169=5 a-12 b-3 \\
& \Rightarrow 13 a^2+13 b^2-83 a+64 b+172=0
\end{aligned}
$
So, the locus of the point is $13 x^2+13 y^2-83 x+64 y+172=0$
Question 3.
Find the Locus of the mid points of the portion of the line $x \cos \theta+\mathrm{y} \sin \theta=p$ intercepted between the axis.
Solution:
Given equation of the line is $\mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathrm{p}$.. (i)
Let $C(h, k)$ be the mid point of the given line $A B$ where it meets the two axis at $A(a, 0)$ and $B(0, b)$.
Since $(a, 0)$ lies on eq (i) then "a $\cos \theta+\theta=p$ "
$
\Rightarrow a=\frac{p}{\cos \theta}
$

$B(0, b)$ also lies on the eq (i) then $0+\mathrm{b} \sin \theta=\mathrm{p}$
$
\Rightarrow b=\frac{p}{\sin \theta}
$
Since $C(h, k)$ is the mid point of $A B$
$
\therefore h=\frac{0+a}{2} \Rightarrow a=2 h \text { and } k=\frac{b+0}{2} \Rightarrow b=2 k
$
Putting the values of $a$ and $b$ is eq (ii) and (iii) we get $P$
$
2 h=\frac{p}{\cos \theta} \Rightarrow \cos \theta=\frac{p}{2 h}
$
and $2 k=\frac{p}{\sin \theta} \Rightarrow \sin \theta=\frac{p}{2 k}$
Squaring and adding eq (iv) and (v) we get
$
\Rightarrow \cos ^2 \theta+\sin ^2 \theta=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \Rightarrow 1=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2}
$
So, the locus of the mid point is $1=\frac{p^2}{4 x^2}+\frac{p^2}{4 y^2}$
$
\Rightarrow 4 x^2 y^2=p^2\left(x^2+y^2\right)
$
Question 4.
Show that the locus of the mid-point of the segment intercepted between the axes of th variable line $x \cos \alpha+y \sin \alpha=p$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$ where $p$ is a constant.

Solution:
The given equation is $x \cos \alpha+y \sin \alpha=p$ or, $\frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}=1$
This cuts the coordinate axes at $\mathrm{A}(p / \cos \alpha, 0)$ and $\mathrm{B}(0, p / \sin \alpha)$. Let $\mathrm{P}(h, k)$ be the $\mathrm{n}$ point of the intercept $\mathrm{AB}$. Then,

$\begin{aligned}
& h=\frac{p / \cos \alpha+0}{2}, k=\frac{0+p / \sin \alpha}{2} \\
& \Rightarrow h=\frac{p}{2 \cos \alpha}, k=\frac{p}{2 \sin \alpha} \\
& \Rightarrow \cos \alpha=\frac{p}{2 h}, \sin \alpha=\frac{p}{2 k} \ldots(i)
\end{aligned}$

Here, $\alpha$ is a variable. To find the locus of $\mathrm{P}(\mathrm{h}, \mathrm{k})$, we have to eliminate $\alpha$. From (i), we obtain
$
\cos ^2 \alpha+\sin ^2 \alpha=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \Rightarrow 1=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}
$
Hence, the locus of $(h, k)$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$
Question 5.
The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$, where $c$ is a constant. Fint the locus of the foot of the perpendicular from the origin on the given line.
Solution:

Let $\mathrm{P}(h, k)$ be the foot of the perpendicular from the origin $\mathrm{O}$ on the line $\frac{x}{a}+\frac{y}{b}=1$ which cuts the coordinates axes at $\mathrm{A}(a, 0)$ and $\mathrm{B}(0, b)$. Then,
Slope of OP $\times$ Slope of AB $=-1$

$
\begin{aligned}
& \Rightarrow \frac{k-0}{h-0} \times \frac{b-0}{0-a}=-1 \\
& \Rightarrow b k=a h \\
& \Rightarrow b=\frac{a h}{k}
\end{aligned}
$
Also $\mathrm{P}(h, k)$ lies on $\frac{x}{a}+\frac{y}{b}=1$
$
\begin{aligned}
& \therefore \frac{h}{a}+\frac{k}{b}=1 \\
& \left.\Rightarrow \frac{h}{a}+\frac{k^2}{a h}=1 \quad \text { [Using }(i)\right] \\
& \Rightarrow a=\frac{h^2+k^2}{h}
\end{aligned}
$

Substituting this values of $a$ in $(i)$, we obtain
$
b=\frac{h^2+k^2}{k}
$
Substituting this values of $a$ and $b$ in $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$, we obtain $\frac{h^2}{\left(h^2+k^2\right)^2}+\frac{k^2}{\left(h^2+k^2\right)^2}=\frac{1}{c^2}$ or $h^2+k^2=c^2$
Hence, the locus of $(h, k)$ is $x^2+y^2=c^2$