6. ⇒ (MHT CET 2021 21th September Evening Shift )

If the lines $x^{2} - 4 x y + y^{2} = 0$ make angles $α$ and $β$ with positive direction X-axis, then $\cot^{2} α + \cot^{2} β =$

A. 14

B. 16

C. 18

D. 20

Correct answer option is (A)

We have lines $x^{2} - 4 x y + y^{2} = 0$ and slopes of the lines are $\tan α$ and $\tan β$ .

$∴ \tan α + \tan β = 4$ and $\tan α \cdot \tan β = 1$

$∴ \tan α = \frac{1}{\tan β} = \cot β$ .

$∴ \cot β + \tan β = 4$ and squaring, we get $\cot^{2} β + \tan^{2} β + 2 = 16 \Rightarrow \tan^{2} β + \cot^{2} β = 14$

7. ⇒ (MHT CET 2021 21th September Morning Shift )

If the two lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ make inclinations $\propto$ and $β$ , then $\tan (α + β) =$

A. $\frac{h}{a + b}$

B. $\frac{2 h}{a + b}$

C. $\frac{h}{a - b}$

D. $\frac{2 h}{a - b}$

Correct answer option is (D)

Lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ make inclinations $α$ and $β$ .

$∴$ $\tan α + \tan β = \frac{- 2 h}{b}$ and $\tan α \tan β = \frac{a}{b}$

Now $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} = \frac{(- \frac{2 h}{b})}{1 - (\frac{a}{b})} = \frac{- 2 h}{b} \times \frac{b}{(b - a)}$

$∴$ $\tan (α + β) = \frac{- 2 h}{b - a} = \frac{2 h}{a - b}$

8. ⇒ (MHT CET 2021 20th September Evening Shift )

If the equation $3 x^{2} - k x y - 3 y^{2} = 0$ represents the bisectors of angles between the lines $x^{2} - 3 x y - 4 y^{2} = 0$ , then value of k is

A. -6

B. -10

C. 6

D. 10

Correct answer option is (B)

We have $x^{2} - 3 x y - 4 y^{2} = 0$ and comparing it with standard equation, we write $A = 1, H = \frac{- 3}{2}, B = - 4$ .

Equation of bisector of angle of this line is

$\begin{aligned} \frac{x^{2} - y^{2}}{A - B} = \frac{x y}{H} \Rightarrow \frac{x^{2} - y^{2}}{1 + 4} = \frac{x y}{(\frac{- 3}{2})} \\ ∴ - 3 x^{2} + 3 y^{2} = 10 x y \Rightarrow 3 x^{2} + 10 x y - 3 y^{2} = 0 \\ Comparing with given equation, we get k = - 10 \end{aligned}$

Comparing with given equation, we get $k = - 10$

9. ⇒ (MHT CET 2021 20th September Evening Shift )

The joint equation of pair of lines through the origin and making an equilateral triangle with the line $y = 3$ is

A. $x^{2} + 3 y^{2} = 0$

B. $3 x^{2} - y^{2} = 0$

C. $x^{2} - 3 y^{2} = 0$

D. $3 x^{2} + y^{2} = 0$

Correct answer option is (B)

Let $△ OAB$ be the required triangle. Since $△ OAB$ is an equilateral triangle.

Slope of line $OA = \tan 60^{\circ} = \sqrt{3}$ and Slope of line $OB = \tan 120^{\circ} = - \sqrt{3}$

$∴$ Equation of $OA$ is $y = \sqrt{3} x$ i.e. $\sqrt{3} x - y = 0$ and equation of OB is $y = - \sqrt{3} x$ i.e. $\sqrt{3} x + y = 0$

Hence required joint equation is

$(\sqrt{3} x - y) (\sqrt{3} x + y) = 0 i.e. 3 x^{2} - y^{2} = 0$

10. ⇒ (MHT CET 2021 20th September Morning Shift )

If the acute angle between the lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ is $\frac{π}{4}$ , then $4 h^{2} =$

A. $(a + 2 b) (a + 3 b)$

B. $a^{2} + 4 a b + b^{2}$

C. $a^{2} + 6 a b + b^{2}$

D. $(a - 2 b) (2 a + b)$

Correct answer option is (C)

As per data given, we write

$\tan \frac{π}{4} = | \frac{2 \sqrt{h^{2} - a b}}{a + b} | = 1$

Squaring both sides, we get

$\begin{aligned} (a + b)^{2} = 4 (h^{2} - a b) \\ ∴ 4 h^{2} = a^{2} + 6 a b + b^{2} \end{aligned}$

⇒Pair of Straight Line