Question

resonalise the denominator of rote over a+x - rote over a-x by rote over a+x - rote over a-x ​

Answer 1

We rationalise the denominator by multiplying both the numerator and the denominator by {√(a + x) + √(a - x)}

So,

y = {√(a + x) + √(a - x)}{√(a + x) + √(a - x)}/{√(a + x) - √(a - x)}/{√(a + x) + √(a - x)}

= {a + x + 2√(a² - x²) + a - x}/(a + x - a + x)

= {2a + 2√(a² - x²)}/(2x)

⇒ xy = a + √(a² - x²)

⇒ xy - a = √(a² - x²)

Now, squaring both sides, we get

(xy - a)² = a² - x²

⇒ x²y² - 2axy + a² = a² - x²

⇒ xy² - 2ay = - x

⇒ xy² - 2ay + x = 0

Answer 2

Answer:

We rationalise the denominator by multiplying both the numerator and the denominator by {√(a + x) + √(a - x)}

So,

y = {√(a + x) + √(a - x)}{√(a + x) + √(a - x)}/{√(a + x) - √(a - x)}/{√(a + x) + √(a - x)}

= {a + x + 2√(a² - x²) + a - x}/(a + x - a + x)

= {2a + 2√(a² - x²)}/(2x)

⇒ xy = a + √(a² - x²)

⇒ xy - a = √(a² - x²)

Now, squaring both sides, we get

(xy - a)² = a² - x²

⇒ x²y² - 2axy + a² = a² - x²

⇒ xy² - 2ay = - x

⇒ xy² - 2ay + x = 0

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