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The coordinates of a moving point P are acosecbcosec(a2(cosecθ+sinθ),b2(cosecθ-sinθ)) where θ is a variabe parameter. Show hat the equation of the locus P is b2x2 – a2y2 = a2b2 - Mathematics

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प्रश्न

The coordinates of a moving point P are `("a"/2 ("cosec" theta + sin theta), "b"/2 ("cosec" theta - sin theta))` where θ is a variabe parameter. Show hat the equation of the locus P is b2x2 – a2y2 = a2b2

बेरीज

उत्तर

Let P(h, k) be the moving point

We are given P = `["a"/2 ("cosec" theta + sin theta), "b"/2 ("cosec"theta - sin theta)]`

⇒ h = `"a"/2 ("cosec theta + sin theta)`

(i.e.) a[cosec θ  + sin θ] = 2h

⇒ cosec θ + sin θ = `(2"h")/"a"`   ......(1)

`"b"/2 ("cosec"theta - sintheta)` = k

⇒ cosec θ – sin θ = `(2"k")/"b"`  ......(2)

(1) + (2) ⇒ 2cosec θ = `(2"h")/"a" + (2"k")/"b"`

(÷ by 2) ⇒ cosec θ = `"h"/"a" + "k"/"b"` ......(3)

(1) – (2) ⇒ 2 sin θ = `(2"h")/"a" - (2"k")/"b"`

(÷ by 2) ⇒ sin θ = `"h"/"a" - "k"/"b"` ......(4)

But cosec θ sin θ = 1

So from (3) and (4)

⇒ `("h"/"a" + "k"/"b") ("h"/"a" - "k"/"b")` = 1

⇒ `"h"^2/"a"^2 - "k"^2/"b"^2` = 1

So the locus of (h, k) is `x^2/"a"^2 - y^2/"b"^2` = 1

(i.e) `("b"^2x^2 - "a"^2y^2)/("a"^2"b"^2)` = 1

⇒ b2x2 – a2y2 = a2b2

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Locus of a Point
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Two Dimensional Analytical Geometry - Exercise 6.1 [पृष्ठ २४३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 6 Two Dimensional Analytical Geometry
Exercise 6.1 | Q 9 | पृष्ठ २४३

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