A calculus problem by Rocco Dalto

Calculus Level pending

The graph of 4 x 2 + 12 x y + 9 y 2 + 8 13 x + 12 13 y − 65 = 0 {\bf 4 x^2 + 12 xy + 9 y^2 + 8 \sqrt{13} x + 12 \sqrt{13} y - 65 = 0 } are two parallel lines.

If one line is represented by a b x + c b y + b = 0 {\bf a\sqrt{b} \:x + c\sqrt{b} \:y + b = 0 } and the other line is represented by a x + c y + d b = 0 , {\bf ax + cy + d\sqrt{b} = 0, } where g c f ( ∣ a ∣ , ∣ b ∣ , ∣ c ∣ , ∣ d ∣ ) = 1 , {\bf gcf(|a|,|b|,|c|,|d|) = 1, }

Find: ∣ a ∣ + ∣ b ∣ + ∣ c ∣ + ∣ d ∣ . {\bf |a| + |b| + |c| + |d|. }


The answer is 23.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Rocco Dalto
Oct 13, 2016

Using the equations of rotation to rotate the x and y axis we have:

x = x ′ c o s θ − y ′ s i n θ {\bf x = x' cos\theta - y' sin\theta }

y = x ′ s i n θ + y ′ c o s θ {\bf y = x' sin\theta + y' cos\theta }

Replacing the equations of rotation in the original equation and simplifying we obtain:

( 4 c o s 2 θ + 12 c o s θ s i n θ + 9 s i n 2 θ ) x ′ 2 + ( 4 s i n 2 θ − 12 c o s θ s i n θ + 9 c o s 2 θ ) y ′ 2 + {\bf (4 cos^2\theta + 12 cos\theta sin\theta + 9 sin^2\theta)\:x'^2 + (4 sin^2\theta - 12 cos\theta sin\theta + 9 cos^2\theta)\:y'^2 + }

( 12 13 s i n θ + 8 13 c o s θ ) x ′ + ( 12 13 c o s θ − 8 13 s i n θ ) y ′ + ( 5 s i n ( 2 θ ) + 12 c o s ( 2 θ ) ) x ′ y ′ {\bf (12\sqrt{13} sin\theta + 8\sqrt{13} cos\theta)\:x' + (12\sqrt{13} cos\theta - 8\sqrt{13} sin\theta) \:y' + (5 sin(2\theta) + 12 cos(2\theta))\:x'y' } − 65 = 0 {\bf - 65 = 0 }

To eliminate the x ′ y ′ {\bf x'y'} term let 5 s i n ( 2 θ ) + 12 c o s ( 2 θ ) = 0 ⟹ {\bf 5 sin(2\theta) + 12 cos(2\theta) = 0 \implies }

t a n ( 2 θ ) = − 12 5 ⟹ {\bf tan(2\theta) = \frac{-12}{5} \implies } 2 t a n θ 1 − t a n 2 θ = − 12 5 ⟹ {\bf \frac{2 tan\theta}{1 - tan^2\theta} = \frac{-12}{5} \implies }

6 t a n 2 θ − 5 t a n θ − 6 = 0 ⟹ t a n θ = 5 + − 13 12 {\bf 6 tan^2\theta - 5 tan\theta - 6 = 0 \implies tan\theta = \frac{5 +- 13}{12} }

Choosing the positive value for t a n θ ⟹ t a n θ = 3 2 ⟹ {\bf tan\theta \implies tan\theta = \frac{3}{2} \implies } c o s θ = 2 13 a n d s i n θ = 3 13 ⟹ {\bf cos\theta = \frac{2}{\sqrt{13}}\: and \:sin\theta = \frac{3}{\sqrt{13}} \implies }

x = 2 13 x ′ − 3 13 y ′ {\bf x = \frac{2}{\sqrt{13}}\: x' - \frac{3}{\sqrt{13}} y' } a n d y = 3 13 x ′ + 2 13 y ′ {\bf \:and \:y = \frac{3}{\sqrt{13}}\:x' + \frac{2}{\sqrt{13}}\:y' }

Replacing the equations of rotation in the original equation and simplifying we obtain:

169 x ′ 2 + 676 x ′ − 845 = 0 ⟹ x ′ 2 + 4 x ′ − 5 = 0 ⟹ ( x ′ + 5 ) ( x ′ − 1 ) = 0 {\bf 169x'^2 + 676x' - 845 = 0 \implies x'^2 + 4x' - 5 = 0 \implies (x' + 5)(x' - 1) = 0 }

∴ {\bf \therefore } we have two parallel lines x ′ = 1 a n d x ′ = − 5 {\bf x' = 1\: and \: x' = -5 } in the x ′ y ′ {\bf x'y' } plane.

To find parallel lines in the x y {\bf xy } plane:

For x ′ = 1 : {\bf x' = 1: }

y ′ = 0 ⟹ ( 2 13 , 3 13 ) {\bf y' = 0 \implies (\frac{2}{\sqrt{13}},\:\frac{3}{\sqrt{13}}) }

y ′ = 1 ⟹ ( − 1 13 , 5 13 ) ⟹ s l o p e m = − 2 3 ⟹ {\bf y' = 1 \implies (-\frac{1}{\sqrt{13}},\:\frac{5}{\sqrt{13}}) \implies \:slope\: m = -\frac{2}{3} \implies } y − 5 13 = − 2 3 ( x + 1 13 ) ⟹ {\bf y - \frac{5}{\sqrt{13}} = -\frac{2}{3}(x + \frac{1}{\sqrt{13}}) \implies } − 2 13 x − 3 13 y + 13 = 0 {\bf -2\sqrt{13}\:x - 3\sqrt{13}\:y + 13 = 0 }

and for x ′ = − 5 : {\bf x' = -5:}

y ′ = 0 ⟹ ( − 10 13 , − 15 13 ) {\bf y' = 0 \implies (-\frac{10}{\sqrt{13}},\:-\frac{15}{\sqrt{13}}) }

y ′ = 1 ⟹ ( − 13 13 , − 13 13 ) ⟹ s l o p e m = − 2 3 ⟹ {\bf y' = 1 \implies (-\frac{13}{\sqrt{13}},\:-\frac{13}{\sqrt{13}}) \implies \:slope\: m = -\frac{2}{3} \implies } − 2 x − 3 y − 5 13 = 0 {\bf -2x - 3y - 5\sqrt{13} = 0 }

∴ {\bf \therefore } The two parallel lines in the x y {\bf xy } plane are:

− 2 13 x − 3 13 y + 13 = 0 {\bf -2\sqrt{13}\:x - 3\sqrt{13}\:y + 13 = 0 }

− 2 x − 3 y − 5 13 = 0 {\bf -2x - 3y - 5\sqrt{13} = 0 }

⟹ ∣ a ∣ + ∣ b ∣ + ∣ c ∣ + ∣ d ∣ = 23. {\bf \implies |a| +|b| + |c| + |d| = 23. }

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Nikola Djuric
Feb 11, 2017

Multiply our equation with √13 and you should easy conclude b=13,d=-5,c=-3,a=-2, so sum is 23

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...