Question

if p be the length of the perpendicular dropped front the point (a,b) on the line x/a+y/b=1,prove that:1/a²+1/b²=1/p².​

Answer 1

Answer:

$\frac{1}{ { {p}^{2} }^{} } = \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} }$

Step-by-step explanation:

$\frac{x}{a} + \frac{y}{b} = 1 \\ \\ p = \frac{ | \frac{a}{a} + \frac{b}{b} - 1 | }{ \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } } } \\ \\ p = \frac{1}{ \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } } } \\ \\ \\ \frac{1}{ { {p}^{2} }^{} } = \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} }$