Question

The differential equation for all the straight lines which are at a unit distance from the origin is
A. $\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1-\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}$
B. $\lgroup~y+x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1+\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}$
C. $\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1+\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}$
D. $\lgroup~y+x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1-\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}$

Answer 1

Hint: In this question, differentiate the equation of lines with a unit distance from the origin with respect to x. Furthermore, make sure to eliminate a by using the method of substitution or elimination. Also, use the necessary formula whenever required during simplification. Now use the differentiated equation and the equation which we got by eliminating a with further simplification to get our required differential equation.

Formula Used: $x\cos~a+y\sin~a=1$
$\dfrac{\text{d}\lgroup\cos~a\rgroup}{\text{d}x}=-\sin~a$
$\dfrac{\text{d}\lgroup\sin~a\rgroup}{\text{d}x}=\cos~a$
$cosec^{2}~a=1+\cot^{2}a$
$\dfrac{1}{\sin~a}=cosec~a$

Complete step by step solution: The equation of lines with a unit distance from the origin is written as
$x\cos~a+y\sin~a=1$ ……..(i)
We know,
$\dfrac{\text{d}\lgroup\cos~a\rgroup}{\text{d}x}=-\sin~a$
$\dfrac{\text{d}\lgroup\sin~a\rgroup}{\text{d}x}=\cos~a$
Thus, when we differentiate with regard to x, we get
$\cos~a+\dfrac{\text{d}y}{\text{d}x}\sin~a=0$ ………..(ii) (Here, $\dfrac{\text{d}x}{\text{d}x}=1$)
Now, we need to eliminate a. It can be done with the help of equations (i) and (ii).
Let's subtract (i) from equation (ii) by multiplying it with x. That is (i)-x (ii), which gives
$x\cos~a+y\sin~a-x\cos~a-x\dfrac{\text{d}y}{\text{d}x}\sin~a=1$
$\Rightarrow\sin~a\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup=1$
We can write by modifying the above equation
$\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup=\dfrac{1}{\sin~a}$.......(iii)
We know,
$\dfrac{1}{\sin~a}=cosec~a$
Let's substitute the above value in (iii),
$\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup=cosec~a$.....(iv)
Also, equation (ii) can be simplified in the following ways,
$\dfrac{\text{d}y}{\text{d}x}\sin~a=-\cos~a$
$\dfrac{\text{d}y}{\text{d}x}=\dfrac{-\cos~a}{\sin~a}$
This implies
$\dfrac{\text{d}y}{\text{d}x}=-\cot~a$ (since, $\dfrac{\cos~a}{\sin~a}=\cot~a$
Squaring both sides of the given equation yields
$\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=\cot^{2}a$........(v)
Squaring both sides of equation (iv),
$\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=cosec^{2}~a$......(vi)
We know,
$cosec^{2}~a=1+\cot^{2}a$
Substituting the above values in (vi), we get
$\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1+\cot^{2}a$
This implies,
$\lgroup~y-x\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}=1+\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup^{2}$ (using equation equation (v))
Therefore, the above equation is our required differential equation.

Option ‘C’ is correct

Note: Students do the mistake of differentiating the base equation two times. It is to be noted that the base equation has only one parameter. So, we need to differentiate only one time.