Skip to content
Exercise 6.5 · Q21

Q.If the lines represented by the equation 6x2+41xy−7y2=06x^2+41xy-7y^2=0 make angles α\alpha and β\beta with the xx-axis, then tan⁡αtan⁡β=\tan\alpha\tan\beta=

(1) −67-\dfrac67
(2) 67\dfrac67
(3) −76-\dfrac76
(4) 76\dfrac76
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
69% · 89/129 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For a homogeneous pair ax2+2hxy+by2=0ax^2+2hxy+by^2=0, the product of slopes is a/ba/b; here a=6,b=−7a=6,b=-7 so tan⁡αtan⁡β=−67\tan\alpha\tan\beta=-\dfrac67.

A second-degree homogeneous equation ax2+2hxy+by2=0ax^2+2hxy+by^2=0 always represents a pair of straight lines through the origin, with slopes m1=tan⁡αm_1=\tan\alpha and m2=tan⁡βm_2=\tan\beta satisfying m1+m2=−2hbm_1+m_2=-\dfrac{2h}{b} and m1m2=abm_1m_2=\dfrac{a}{b} (obtained by dividing the equation by y2y^2 and treating it as a quadratic in m=x/ym=x/y).

Step 1. Identify a,h,ba,h,b. Comparing 6x2+41xy−7y2=06x^2+41xy-7y^2=0 with ax2+2hxy+by2=0ax^2+2hxy+by^2=0:

a=6,2h=41 (so h=412),b=−7a=6,\quad 2h=41 \ (\text{so } h=\tfrac{41}{2}),\quad b=-7

Step 2. Verify the equation genuinely factors into two real lines. Discriminant condition: h2−ab=(412)2−6(−7)=16814+42>0h^2-ab=\left(\tfrac{41}{2}\right)^2-6(-7)=\tfrac{1681}{4}+42>0, so two distinct real lines exist (as required for α,β\alpha,\beta to be defined). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.