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If x and y are connected parametrically by the equations, without eliminating the parameter, find dy/dx. x = a sec θ, y = b tan θ - Mathematics

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Question

If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a sec θ, y = b tan θ

Sum
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Solution

Here x = a sec θ  ...(1)

y = b tan θ  ....(2)

Differentiating (1) and (2) w.r.t. θ, we get

`dx/(d θ)` = a sec θ tan θ and `dy/(d θ)` = b sec2 θ

`therefore dy/dx = (dy/(d θ))/(dx/(d θ))`

= `(b sec^2 θ)/(a sec θ tan θ)`

= `(b sec θ)/(a tan θ)`

= `b/a` sec θ cot θ

= `b/a xx 1/(cos θ) xx (cos θ)/(sin θ)`

= `b/a xx 1/(sin θ)`

= `b/a` cosec θ

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Chapter 5: Continuity and Differentiability - Exercise 5.6 [Page 181]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.6 | Q 9 | Page 181

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