Question

Find the loci of the following:

$$(i){\text{ }}r = a; \\\ (ii){\text{ }}r = a\sin \theta ; \\\ (iii){\text{ }}r = a\cos \theta ; \\\ (iv){\text{ }}r = a\sec \theta ; \\\ (v){\text{ }}r = a\cos (\theta - \alpha ); \\\ (vi){\text{ }}r = a\sec (\theta - \alpha ); \\\$$

Answer 1

As we all know that locus is a curve formed by joining all the points
satisfying the given relation, let’s use these equations $x = r\cos \theta$, $y = r\sin \theta$ and ${x^2} + {y^2} = {r^2}$ to find the solutions.

Complete step-by-step answer:

So, we are given with some relations and had to find loci of that,

As, we know that for finding the loci we take, two conditions and that were,
$\Rightarrow x = r\cos \theta$ (1)
$\Rightarrow y = r\sin \theta$ (2)
So, ${x^2} + {y^2} = {r^2}$ (3)

So, by using equation 1, 2 and 3 we can find loci.
For finding loci,
$(i){\text{ }}$
It is given that,
$\Rightarrow r = a$ (4)

Now, squaring equation 4 we get,
$\Rightarrow {r^2} = {a^2}$ (5)
$\Rightarrow$Hence, ${x^2} + {y^2} = {r^2}$ (from equation 3 and 5)
$(ii){\text{ }}$
It is given that,
$\Rightarrow r = a\sin \theta$ (6)

Now, putting the value of $\sin \theta$ from equation 2 to equation 6 and then solving. We get,
$\Rightarrow {r^2} = ay$ (7)
\Rightarrow $$$${x^2} + {y^2} = ay (from equation 3 and equation 7)

Now, to solve the above equation. We add $\dfrac{{{a^2}}}{4} - ay$ to both sides of the above equation. We get,
$\Rightarrow {x^2} + {y^2} + \dfrac{{{a^2}}}{4} - ay = {\left( {\dfrac{a}{2}} \right)^2}$
$\Rightarrow$Hence, ${x^2} + {\left( {y - \dfrac{a}{2}} \right)^2} = {\left( {\dfrac{a}{2}} \right)^2}$ (from above equation)
$(iii){\text{ }}$
It is given that,
$\Rightarrow r = a\cos \theta$ (8)

Now, putting the value of $\cos \theta$ from equation 1 to equation 8 and then solving. We get,
$\Rightarrow {r^2} = ax$ (9)
\Rightarrow $$$${x^2} + {y^2} = ax (from equation 3 and equation 9)

Now, to solve the above equation. We add $\dfrac{{{a^2}}}{4} - ax$ to both sides of the above equation. We get,
$\Rightarrow {x^2} + {y^2} + \dfrac{{{a^2}}}{4} - ax = {\left( {\dfrac{a}{2}} \right)^2}$
$\Rightarrow$Hence, ${\left( {x - \dfrac{a}{2}} \right)^2} + {y^2} = {\left( {\dfrac{a}{2}} \right)^2}$ (from above equation)

$(iv)$
It is given that,
$\Rightarrow r = \dfrac{a}{{\cos \theta }}$ (10)
Now, putting the value of $\cos \theta$ from equation 1 to equation 10 and then solving.

We get,
Hence, $x = a$ (from equation 3 and equation 10)
$(v)$

It is given that,
$\Rightarrow r = a\cos \left( {\theta - \alpha } \right)$ (11)

Solving equation 11. We get,
$\Rightarrow r = a\cos \theta \cos \alpha + a\sin \theta {\text{sin}}\alpha$ (12)

Now, putting the value of $\cos \theta$ and $\sin \theta$ from equation 1 and 2 to equation 12 and then solving.
$\Rightarrow {r^2} = ax\cos \alpha + ay\sin \alpha$ (13)
$\Rightarrow$Hence, ${\left( {x - \dfrac{a}{2}\cos \alpha } \right)^2} + {\left( {y - \dfrac{a}{2}\sin \alpha } \right)^2} = {\left( {\dfrac{a}{2}} \right)^2}$ (from equation 3 and equation 13)
$(vi)$

It is given that,
$\Rightarrow r = \dfrac{a}{{\cos \left( {\theta - \alpha } \right)}}$ (14)

Solving equation 14. We get,
$\Rightarrow r = \dfrac{a}{{\cos \theta \cos \alpha + \sin \theta {\text{sin}}\alpha }}$ (15)

Now, putting the value of $\cos \theta$ and $\sin \theta$ from equation 1 and 2 to equation 15 and then solving.
$\Rightarrow$Hence, $x\cos \alpha + y\sin \alpha = a$ (from equation 15)

Note: Whenever we came up with this type of problem where we are asked to find the locus of the given relation then easiest and efficient way to get locus is by reducing the given relation in terms of x, y and a using equations $x = r\cos \theta$, $y = r\sin \theta$ and ${x^2} + {y^2} = {r^2}$