Straight lines JEE

Geometry Level 2

Given the equation 4 x 2 + 2 3 x y + 2 y 2 = 1 4x^2+2\sqrt{3}xy+2y^2=1 .
Through what angles θ ∈ [ 0 , π ] \theta \in [0,\pi] should the axes be rotated so that the term x y xy is removed from the transformed equation?

Give the answer as the sum of all values of θ \theta (in degrees).

90 120 180 105 150

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Ravi Dwivedi
Aug 10, 2015

Suppose we rotate our coordinate axes by angle θ \theta such that our ( x , y ) (x,y) point is now ( x ′ , y ′ ) (x',y') point

We need to draw a figure for the relations I got as

x = x ′ cos ⁡ θ − y ′ sin ⁡ θ x=x' \cos \theta - y' \sin \theta

and y = x ′ sin ⁡ θ + y ′ cos ⁡ θ y=x' \sin \theta + y' \cos \theta

⟹ x ′ = x cos ⁡ θ + y sin ⁡ θ \implies x'=x \cos \theta+ y\sin \theta

and y ′ = − x sin ⁡ θ + y cos ⁡ θ y'=-x \sin \theta + y \cos \theta

Our equation for the curve is now

4 x ′ 2 + 2 3 x ′ y ′ + 2 y ′ 2 = 1 4x'^2 + 2\sqrt{3}x'y'+ 2y'^2=1

⟹ 4 ( x cos ⁡ θ + y sin ⁡ θ ) 2 + 2 3 ( x cos ⁡ θ + y sin ⁡ θ ) ( − x sin ⁡ θ + y cos ⁡ θ ) + 2 ( − x sin ⁡ θ + y cos ⁡ θ \implies 4(x \cos \theta+ y\sin \theta)^2+2\sqrt{3}(x \cos \theta+ y\sin \theta)(-x \sin \theta + y \cos \theta)+2(-x \sin \theta + y \cos \theta =1

As we need to remove the x − y x-y term

Put coefficient of x y = 0 xy=0

2 3 ( cos ⁡ 2 θ − sin ⁡ 2 θ ) + 2 ( 2 − 4 ) sin ⁡ θ cos ⁡ θ = 0 2\sqrt{3}(\cos ^2 \theta - \sin ^2 \theta)+2(2-4)\sin \theta \cos \theta=0

c o t 2 θ = 1 3 cot 2\theta= \frac{1}{\sqrt{3}} which gives θ = 3 0 ∘ \theta=30^{\circ} and θ = 12 0 ∘ \theta = 120^{\circ} in the required interval

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

More generally, to remove that x y xy term, what angle should we rotate by?

In response to challenge master

In general let the curve be A x 2 + B x y + C x 2 + D x + E y + F = 0 Ax^2+Bxy+Cx^2+Dx+Ey+F=0

We rotate the curve by θ \theta to remove the x − y x-y term.

And cot ⁡ 2 θ = A − C B \cot 2\theta = \frac{A-C}{B}

Proof is same as the working above

Ravi Dwivedi - 5 years, 10 months ago

i clicked the option 120 nd it still shows wrong .

A Former Brilliant Member - 4 years, 5 months ago

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Same here 😶

Sabhrant Sachan - 4 years, 5 months ago

The question asks for " sum of all values of theta..." Given the above solution, there are 2 values of theta which satisfy the requirement that theta is in the required interval. One of them is 120 degrees, but the sum is 150 degrees.

Tom Capizzi - 4 years, 5 months ago

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