Manipal Entrance Test 2019 Math Paper

Let θ be the angle between the given pair of straight lines
x2−3xy+λy2 +3x−5y+2=0
∴θ=tan−1

1

3

(given)
or tan‌θ=

1

3

We know that,
tan‌θ=

2√h2−ab

a+b

where, h=−

3

2

,a=1,b=λ
put these value in Eq. (ii), we get

1

3

=

2√ 9 4 −λ.1

1+λ

⇒(1+λ)=3√9−4λ
On squaring both sides, we get
(1+λ)2=(3√9−4λ)2
⇒1+λ2+2λ=81−36λ
⇒λ2+38λ−80=0
⇒λ2+38λ−80=0
⇒λ2+40λ−2λ−80=0
⇒λ(λ+40)−2(λ+40)=0
⇒(λ−2)(λ+40)=0
⇒λ=2,−40⇒λ⇒=2
(∵ λ is non-negative real number))