Question

please help for points

Answer 1

$\textbf{Answer :}$

Given that,

x = a sinθ ...(i)

y = b cosθ ...(ii)

Now, differentiating both sides of (i) and (ii) with respect to θ, we get

dx/dθ = a cosθ ...(iii)

dy/dθ = - b sinθ ...(iv)

So, dy/dx

= (dy/dθ)/(dx/dθ)

= (- b sinθ)/(a cosθ)

= - b/a tanθ

Now, differentiating with respect to x, we get

d²y/dx²

= - b/a sec²θ (dθ/dx)

= - b/a sec²θ (1/a cosθ), by (iii)

= - b/a² sec³θ

Therefore, $\textbf{Option (C)}$ is correct.

#$\bold{MarkAsBrainliest}$

Answer 2

GIVEN

• x = a sinθ _(i)
• y = b cosθ _(ii)

Differentiate both eqn (i) and (ii) with respect to θ

• dx/dθ = a cosθ _(iii)
• dy/dθ = - b sinθ _(iv)

NOW

• dy/dx
• (dy/dθ) / (dx/dθ)
• (- b sinθ)/(a cosθ)
• - b/a tanθ

Differentiating with respect to x, we get

• d²y/dx²
• - b/a sec²θ (dθ/dx)
• - b/a sec²θ (1/a cosθ) [ dx/dθ = a cosθ ]
• - b/a² sec³θ