ABC is a right angled triangle in which `/_C = 90^@`. If `BC = a, AB = c ,CA = b `and the length of perpendicular from C to AB be p. then, `1/(a^2) + 1/(b^2) = ?`

Options

A. `1/p`
B. `2/(p^2)`
C. `1/(p^2)`
D. None of these

Answer

To solve the problem step by step, we will use the properties of right-angled triangles and the relationship between the sides and the area.

### Step-by-Step Solution:

1. **Understand the Triangle**: We have a right-angled triangle ABC where angle C is 90 degrees. The sides are defined as follows:
- BC = a (opposite to angle A)
- CA = b (opposite to angle B)
- AB = c (the hypotenuse)

2. **Area of the Triangle**: The area of triangle ABC can be calculated using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
If we take BC (a) as the base and CA (b) as the height, the area can be expressed as:
\[ \text{Area} = \frac{1}{2} \times a \times b \]

3. **Using the Perpendicular**: The length of the perpendicular from point C to line AB is given as p. The area can also be expressed using this perpendicular:
\[ \text{Area} = \frac{1}{2} \times c \times p \]
where c is the length of the hypotenuse AB.

4. **Equating the Areas**: Since both expressions represent the area of the same triangle, we can set them equal to each other:
\[ \frac{1}{2} \times a \times b = \frac{1}{2} \times c \times p \]
Simplifying this gives:
\[ a \times b = c \times p \]

5. **Squaring the Equation**: We will square both sides of the equation:
\[ (a \times b)^2 = (c \times p)^2 \]
This expands to:
\[ a^2 \times b^2 = c^2 \times p^2 \]

6. **Using Pythagoras' Theorem**: From the Pythagorean theorem, we know that:
\[ c^2 = a^2 + b^2 \]
Substituting this into our equation gives:
\[ a^2 \times b^2 = (a^2 + b^2) \times p^2 \]

7. **Rearranging the Equation**: Rearranging this equation, we can express it in terms of \( \frac{1}{a^2} + \frac{1}{b^2} \):
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{p^2}{a^2 \times b^2} \]

8. **Final Expression**: Therefore, we can conclude that:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{p^2}{a^2 \times b^2} \]

### Final Answer:
Thus, the value of \( \frac{1}{a^2} + \frac{1}{b^2} \) is \( \frac{p^2}{a^2 \times b^2} \).