Question

`\underline{\boxed{\blue{\Large{\bf{Challenge}}}}}`

If 'P and 'Q' are two points whose coordinates are (at^2, 2at) and (a/t^2, 2a/t) respectively and S is the point(0, 0). Show that 1/SP + 1/SQ is independent of 't'.

Note:-

Plagarised/spam/short answers will be deleted on the spot.

Answer with all steps and proper explanation .
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Answer 1

Explanation:

Given point are: P(at², 2at)

Q(a/t², 2a/t)

S(0, 0)

Now, the distance between A(x₁y₁) and B(X₂, y₂)

then AB = √{(x₂ - x₁)² + (y₂ - y₁)²} units

(i) The distance between S and P:

(x₁, y₁) = (0, 0) ⇛x₁= 0, y₁ = 0

(x₂, y₂ ) = (at², 2at) ⇛x₂ = at², y₂ = 2at

SP = √{at² - 0)² + (2at - 0)²}

= √{(at²)² + (2at)²}

= √{a²t²*² + 4a²t²}

= √{a²t⁴ + 4a²t²}

= √{a²t²(t²+4)}

SP = at√(t² + 4)→→→Eqn(1)

(ii) The distance between S and Q :

(x₁, y₁) = (0, 0) ⇛x₁= 0, y₁ = 0

(x₂, y₂ ) = (a/t², 2a/t) ⇛x₂ = a/t², y₂ = 2a/t

SQ = √[{(a/t²) - 0} + {(2a/t) - 0}

= √{(a/t²)² + (2a/t)²}

= √{(a²/t²*²) + (4a²/t²)}

= √{(a²/t⁴) + (4a²/t²)}

= √{(a² + 4a²t²)/t⁴}

= √[{a²(1 + 4t²)}/t⁴]

SQ = (a/t²)√(1 + 4t²) →→→ Eqn(2)

Now,

(1/SP) + (1/SQ) = [1/{at√(t² + 4)}] + [1/{(a/t²)√(1 + 4t²)}]

= (1/at)[1/{√(t² + 4)}] + (t²/a)[1/{(√1 + 4t²)}]

= (1/a)[[1/{t√(t² + 4)}] + [t²/{√(1 + 4t²)}]]

(1/SP) + (1/SQ) = 1/a is not independent of 't'

If suppose S = (a, 0) then

SP = √{(at² - a)² + (2at - 0)²}

= √{a²(t² - 1)² + (2at)²}

= a√{(t² - 1)² + 4t²}

= a√{(t² + 1)²}

SP = a(t² + 1)

1/SP = 1/{a(t² + 1)} →→→Eqn(1)

And

SQ = √[{(a/t²) - a}² + {(2a/t) - 0}²]

= √[a²{(1/t²) - 1}² + a²(2/t)²]

= a√[{(1 - t²)²/t⁴} + (4/t²)]

= a√[{(1 - t²)² + 4t²}/t⁴]

( a/t²)√(1 + t²)²

SQ = (a/t²)(1 + t²)

1/SQ = 1/{(a/t²)(1 + t²)} = t²/{a(1 + t²)} →→→Eqn(2)

Therefore, (1/SP) + (1/SQ)

= 1/{a(t² + 1)} + t²/{a(1 + t²)}

= (1 + t²)/a(1 + t²)

= 1/a

1/a is independent of 't'.