Name: Finding the Angle Between Two Following Curves
Uploaded: 2019-07-19
Duration: 11 min 3 s
Description: In this video, you will learn how to find the angle between two following curves, through a simple method. Watch this video and learn more now.

Question

Find the equation of the tangent and the normal to the following curve at the indicated point

c^{2} (x^{2} + y^{2}) = x^{2} y^{2}

at

(\frac{c}{cosθ}, \frac{c}{sinθ})

?

Accepted Answer

Given equation of curve is $c^{2} (x^{2} + y^{2}) = x^{2} y^{2}$

Differentiating both sides with respect to $x$

$\Rightarrow 2 {xc}^{2} + 2 {yc}^{2} \frac{dy}{dx} = x^{2} (2 y \frac{dy}{dx}) + 2 {xy}^{2}$

$\Rightarrow \frac{dy}{dx} (2 {yc}^{2} - 2 x^{2} y) = 2 {xy}^{2} - 2 {xc}^{2}$

$\Rightarrow \frac{dy}{dx} = \frac{{xy}^{2} - {xc}^{2}}{{yc}^{2} - x^{2} y}$

Slope of tangent, $m = {(\frac{dy}{dx})}_{(\frac{c}{cosθ}, \frac{c}{sinθ})}$

$= \frac{\frac{c^{3}}{{cosθsin}^{2} θ} - \frac{c^{3}}{cosθ}}{\frac{c^{3}}{sinθ} - \frac{c^{3}}{\cos^{2} θsinθ}}$

$= \frac{\frac{1 - \sin^{2} θ}{{cosθsin}^{2} θ}}{\frac{\cos^{2} θ - 1}{\cos^{2} θsinθ}}$

$= \frac{\cos^{2} θ}{{cosθsin}^{2} θ} \times \frac{\cos^{2} θsinθ}{- \sin^{2} θ}$

$= \frac{- \cos^{3} θ}{\sin^{3} θ}$

Given $(x_{1}, y_{1}) = (\frac{c}{cosθ}, \frac{c}{sinθ})$

Equation of tangent is $y - y_{1} = m (x - x_{1})$

$\Rightarrow y - \frac{c}{sinθ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (x - \frac{c}{cosθ})$

$\Rightarrow \frac{ysinθ - c}{sinθ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (\frac{xcosθ - c}{cosθ})$

$\Rightarrow \sin^{2} θ (ysinθ - c) = - \cos^{2} θ (xcosθ - c)$

$\Rightarrow {ysin}^{3} θ - {csin}^{2} θ = - {xcos}^{3} θ + {ccos}^{2} θ$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ = c (\sin^{2} θ + \cos^{2} θ)$

$\Rightarrow {xcos}^{3} θ + {ysin}^{3} θ = c$

Equation of normal is $y - y_{1} = \frac{- 1}{m} (x - x_{1})$

$\Rightarrow y - \frac{c}{sinθ} = \frac{\sin^{3} θ}{\cos^{3} θ} (x - \frac{c}{cosθ})$

$\Rightarrow \cos^{3} θ (y - \frac{c}{sinθ}) = \sin^{3} θ (x - \frac{c}{cosθ})$

$\Rightarrow {ycos}^{3} θ - \frac{{ccos}^{3} θ}{sinθ} = {xsin}^{3} θ - \frac{{csin}^{3} θ}{cosθ}$

$\Rightarrow {xsin}^{3} θ - {ycos}^{3} θ = \frac{{csin}^{3} θ}{cosθ} - \frac{{ccos}^{3} θ}{sinθ}$

$\Rightarrow {xsin}^{3} θ - {ycos}^{3} θ = c (\frac{\sin^{4} θ - \cos^{4} θ}{cosθsinθ})$

$\Rightarrow {xsin}^{3} θ - {ycos}^{3} θ = c [\frac{(\sin^{2} θ + \cos^{2} θ) (\sin^{2} θ - \cos^{2} θ)}{cosθsinθ}]$

$\Rightarrow \sin^{3} θ - {ycos}^{3} θ = 2 c [\frac{- (\cos^{2} θ - \sin^{2} θ)}{2 cosθsinθ}]$

$\Rightarrow \sin^{3} θ - {ycos}^{3} θ = 2 c [\frac{- \cos (2 θ)}{\sin (2 θ)}]$

$\Rightarrow \sin^{3} θ - {ycos}^{3} θ = - 2 ccot (2 θ)$

$\Rightarrow \sin^{3} θ - {ycos}^{3} θ + 2 ccot (2 θ) = 0$

Find the equation of the tangent and the normal to the following curve at the indicated point $c^{2} (x^{2} + y^{2}) = x^{2} y^{2}$ at $(\frac{c}{cosθ}, \frac{c}{sinθ})$ ?

Important Questions on Tangents and Normals

Learn and Prepare for any exam you want !

Find the equation of the tangent and the normal to the following curve at the indicated point c2x2+y2=x2y2 at ccosθ,csinθ ?

Important Questions on Tangents and Normals

Learn and Prepare for any exam you want !

Find the equation of the tangent and the normal to the following curve at the indicated point $c^{2} (x^{2} + y^{2}) = x^{2} y^{2}$ at $(\frac{c}{cosθ}, \frac{c}{sinθ})$ ?