Question

If the differential equation of all straight lines which are at a fixed distance of $10$ units from origin is
$(y-x{y}_1{)}^2=A(1+y_1^2)$ then $\frac{A}{100}$ is equal to ... units.

Answer 1

The given family of lines can be represented by the equation
$x\cos \alpha +y\sin \alpha =10$ .......... $\left(1\right)$
where $\alpha$ is an arbitrary constant. Differentiating we have
$\cos \alpha +\sin \alpha \frac{dy}{dx}=0$ ...... $\left(2\right)$
Multiplying (2) by $x$ and subtracting it from (1), we get
$y\sin \alpha -x\sin \alpha \frac{dy}{dx}=10$
$\Rightarrow (y-x{y}_1)\sin \alpha =10$ ...... $\left(3\right)$
Multiplying (1) by $y_1=\frac{dy}{dx}$ and (2) by $y$ and then subtracting, we get
$x{y}_1\cos \alpha -y\cos \alpha =10y_1$
$\Rightarrow (x{y}_1-y)\cos \alpha =10y_1$ ...... $\left(4\right)$
Squaring $\left(3\right)$ and $\left(4\right)$ and then adding, we get

$(y-x{y}_1{)}^2=100(1+y_1^2)$

Comparing with the given equation $(y-x{y}_1{)}^2=A(1+y_1^2)$, we get

$A=100$

$\Rightarrow \frac{A}{100}=1$