Let's start by examining the given equation of the line and the point from which the perpendicular distance is measured.
The equation provided for the line is:
[tex]\[ \frac{x}{a} + \frac{y}{b} = 1 \][/tex]
To find the perpendicular distance [tex]\( p \)[/tex] from the point [tex]\( (a, b) \)[/tex] to this line, we use the formula for the distance from a point to a line in the form [tex]\( Ax + By + C = 0 \)[/tex]. However, it's more straightforward to use the distance formula directly adapted for a line given in intercept form.
In intercept form, the line equation [tex]\( \frac{x}{a} + \frac{y}{b} = 1 \)[/tex] intersects the axes at [tex]\( (a, 0) \)[/tex] and [tex]\( (0, b) \)[/tex].
The general formula for the distance from a point [tex]\( (x_0, y_0) \)[/tex] to the line [tex]\(\frac{x}{a} + \frac{y}{b} = 1 \)[/tex] can be given as:
[tex]\[ p = \frac{| \frac{x_0}{a} + \frac{y_0}{b} - 1 |}{\sqrt{\left( \frac{1}{a} \right)^2 + \left( \frac{1}{b} \right)^2}} \][/tex]
For our specific problem, the point is [tex]\( (a, b) \)[/tex].
Plugging in [tex]\( a \)[/tex] for [tex]\( x_0 \)[/tex] and [tex]\( b \)[/tex] for [tex]\( y_0 \)[/tex]:
[tex]\[ p = \frac{| \frac{a}{a} + \frac{b}{b} - 1 |}{\sqrt{\left( \frac{1}{a} \right)^2 + \left( \frac{1}{b} \right)^2}} \][/tex]
[tex]\[ p = \frac{| 1 + 1 - 1 |}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \][/tex]
[tex]\[ p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \][/tex]
Given that the distance is [tex]\( p \)[/tex], we now solve for [tex]\( \frac{1}{a^2} + \frac{1}{b^2} \)[/tex].
Squaring both sides of the equation:
[tex]\[ p^2 = \left( \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right)^2 \][/tex]
[tex]\[ p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} \][/tex]
Therefore, inverting both sides to isolate [tex]\( \frac{1}{a^2} + \frac{1}{b^2} \)[/tex]:
[tex]\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} \][/tex]
Hence, the required expression for [tex]\( \frac{1}{a^2} + \frac{1}{b^2} \)[/tex] is:
[tex]\[ \boxed{\frac{1}{p^2}} \][/tex]
