Question

Find the condition that the line $x\cos \alpha +y\sin \alpha =p$ may touch the curve:

$\frac{x^m}{a^m}+\frac{y^m}{b^m}=1$
Then the condition is ${\left(acos\alpha \right)}^{m/(m-1)}+{\left(bsin\alpha \right)}^{m/(m-1)}=p^{m/(m-1)}$
if true enter 1 else enter 0

Answer 1

The tangent to the given curveat any point $(x,y)$ is

$\frac{X}{a}{\left(\frac{x}{a}\right)}^{m-1}+\frac{Y}{b}{\left(\frac{y}{b}\right)}^{m-1}=1$ ...(1)
Compare with $X\cos a+Y\sin a=p$ ...(2)

$\frac{{\left(x/a\right)}^{m-1}}{acos\alpha }=\frac{{\left(y/b\right)}^{m-1}}{bsin\alpha }=\frac{1}{p}$
$\therefore \frac{x}{a}={\left(\frac{acos\alpha }{p}\right)}^{1/(m-1)},$
$\therefore \frac{y}{b}={\left(\frac{bsin\alpha }{p}\right)}^{1/(m-1)},$ ...(3)
But the point $(x,y)$ lies on ${\left(\frac{x}{a}\right)}^m+{\left(\frac{y}{b}\right)}^m=1.$ ....(4)
Putting for $\frac{x}{a}$ and $\frac{y}{b}$ from (3) in (4), we get

${\left(\frac{acos\alpha }{p}\right)}^{m/(m-1)}+{\left(\frac{bsina}{p}\right)}^{m/(m-1)}=1$
$\Rightarrow {\left(acos\alpha \right)}^{m/(m-1)}+{\left(bsin\alpha \right)}^{m/(m-1)}=p^{m/(m-1)}$