Question

If q₁ and q₂ be the angles which the lines (x² + y²) (cos²q sin²a + sin²q) = (x tan a – y sin q)² make with the axis of x,

and q = $\frac{\pi}{6}$, then tan q₁ + tan q₂ is equal to –

• A. – $\frac{8}{3}$sina
• B. – $\frac{8}{3}$cosec 2a
• C. – $8\sqrt{3}$cosec 2a
• D. –4cosec 2a

Answer 1

Answer: (B)

– $\frac{8}{3}$cosec 2a

Answer 2

The given equation can be written as

(x² + y²) (cos² q sin² a + sin² q) = x² tan²

a – 2xy tan a sin q + y²sin² q

or (cos² q sin² a + sin² q – tan² a)x² + 2(tan a sin q) xy + cos² q sin² a y² = 0

Since the slope of these lines are given as tan q₁ and tan q₂

Sum of the slopes = $\frac { - 2 \tan \alpha \sin \theta } { \cos ^ { 2 } \theta \sin ^ { 2 } \alpha }$ $\left( \because\theta = \frac{\pi}{6} \right)$

Ž tan q₁ + tan q₂ = $\frac{- 2\tan\alpha \times \frac{1}{2}}{\frac{3}{4} \times \sin^{2}\alpha}$= $\frac{8}{3}$cosec2a.