Write the area of the triangle formed by the coordinate axes and the line $s e c θ - t a n θ) x + (s e c θ + t a n θ) y = 2.$

Open in App

Solution

$(s e c θ - t a n θ) x + (s e c θ + t a n θ) y = 2$

$\frac{x}{\frac{2}{(s i n θ - t a n θ)}} + \frac{y}{\frac{2}{(s e c θ + t a n θ)}} = 1$

$S o, a = \frac{2}{s e c θ - t a n θ}, b = \frac{2}{s e c θ + t a n θ}$

Area of required triangle

$A = \frac{1}{2} a b$

$= \frac{1}{2} (\frac{2}{s e c θ - t a n θ}) (\frac{2}{s e c θ + t a n θ})$

$= \frac{2}{s e c^{2} θ - t a n^{2} θ}$

A=2 sq,units

Suggest Corrections

Similar questions

x (tan θ) + y (cot θ) = 4

θ \in (0, \frac{π}{2}),

θ

x sin θ + y cos θ = 4

\frac{x^{2}}{27} + y^{2} = 1

3 \sqrt{3} c o s θ, s i n θ

0 < θ < \frac{π}{2}

θ

y = 2 x + 4

a x + b y = 2 a b

Join BYJU'S Learning Program