Q.If $p$ is the length of perpendicular from the origin on the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $a^2$, $p^2$, $b^2$ are in A.P, then show that $a^4+b^4=0$.

The problem involves two core concepts: finding the perpendicular distance from a point to a line, and the property of terms in an Arithmetic Progression (A.P.).

First, we express the given line equation in the standard form $Ax+By+C=0$. The line $\dfrac{x}{a}+\dfrac{y}{b}=1$ can be rewritten as $bx+ay=ab$, or $bx+ay-ab=0$.

Next, we use the formula for the perpendicular distance $p$ from the origin $(0,0)$ to the line $bx+ay-ab=0$:

Applying this, $p = \dfrac{|b(0)+a(0)-ab|}{\sqrt{b^2+a^2}} = \dfrac{|-ab|}{\sqrt{a^2+b^2}} = \dfrac{ab}{\sqrt{a^2+b^2}}$.

Squaring both sides gives $p^2 = \dfrac{a^2b^2}{a^2+b^2}$. …

To solve this problem, we need to combine two key concepts from coordinate geometry and sequences: the formula for the perpendicular distance from a point to a line, and the property of terms in an Arithmetic Progression (AP).

Concept and Intuition

1. Perpendicular Distance from a Point to a Line:

   The perpendicular distance is the shortest distance from a given point to a line. Imagine dropping a plumb line from the point straight down to the line; the length of that plumb line is the perpendicular distance. For a general line $Ax + By + C = 0$ and a point $(x_1, y_1)$, the distance $p$ is given by a specific formula. When the point is the origin $(0,0)$, this formula simplifies nicely. The absolute value ensures that distance, being a length, is always non-negative.

2. Arithmetic Progression (AP):

   An Arithmetic Progression is a sequence of numbers where the difference between consecutive terms is constant. For example, $2, 5, 8, 11, \dots$ is an AP with a common difference of $3$. A fundamental property is that if three terms $X, Y, Z$ are in AP, then the middle term $Y$ is the arithmetic mean of the other two, meaning $Y = \dfrac{X+Z}{2}$, or equivalently, $2Y = X+Z$. We will use this property to relate $a^2, p^2,$ and $b^2$.

Let's apply these ideas step-by-step.

Step-by-Step Solution

1. Rewrite the line equation in the general form $Ax + By + C = 0$.

   The given equation of the line is $\dfrac{x}{a}+\dfrac{y}{b}=1$.

   To remove the denominators, we multiply the entire equation by $ab$:

   $b x + a y = ab$

   Now, rearrange it into the standard general form $Ax + By + C = 0$:

   $b x + a y - ab = 0$

   Here, $A=b$, $B=a$, and $C=-ab$.

2. Calculate the perpendicular distance $p$ from the origin $(0,0)$ to the line.

   The perpendicular distance $p$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by:

   $p = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

   For the origin $(0,0)$, we substitute $x_1=0$ and $y_1=0$:

   $p = \dfrac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}}$

   $p = \dfrac{|-ab|}{\sqrt{a^2 + b^2}}$

   Since $p$ represents a length, it must be positive. We can write $|-ab| = |ab|$.

   So, $p = \dfrac{|ab|}{\sqrt{a^2 + b^2}}$.

   To make calculations easier, we can square both sides to get rid of the absolute value and the square root: …