Prove that the angle between the straight lines joining the origin to the intersection of the straight line $y = 3 x + 2$ with the curve $x^{2} + 2 x y + 3 y^{2} + 4 x + 8 y - 11 = 0$ is ${tan}^{- 1} \frac{2 \sqrt{2}}{3}$

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Solution

$x^{2} + 2 x y + 3 y^{2} + 4 x + 8 y - 11 = 0 y = 3 x + 2$

The equation of pair straight line joining the origin is find by method of homogenisation. We make the coefficient of constant term of straight line $1$ and put it the equation of the curve so that degree of each term of the curve become $2$

$\frac{y - 3 x}{2} = 1... (i) x^{2} + 2 x y + 3 y^{2} + 4 x (1) + 8 y (1) - 11 {(1)}^{2} = 0 x^{2} + 2 x y + 3 y^{2} + 4 x (\frac{y - 3 x}{2}) + 8 y (\frac{y - 3 x}{2}) - 11 {(\frac{y - 3 x}{2})}^{2} = 0 x^{2} + 2 x y + 3 y^{2} + 2 x y - 6 x^{2} + 4 y^{2} - 12 x y - \frac{11 y^{2}}{4} - \frac{99 x^{2}}{4} + \frac{33 x y}{2} = 0 - 119 x^{2} + 17 y^{2} + 34 x y = 0 7 x^{2} - y^{2} - 2 x y = 0$

is the desired equation of straight lines.

Angle between pair of straight lines that is $tan θ = ∣ ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a + b} ∣ ∣ ∣ ∣$

$tan θ = ∣ ∣ ∣ \frac{2 \sqrt{1 + 7}}{7 - 1} ∣ ∣ ∣ = \frac{2 \sqrt{2}}{3} \Rightarrow θ = {tan}^{- 1} (\frac{2 \sqrt{2}}{3})$

Hence proved.

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Prove that the angle between the straight lines joining the origin to the intersection of the straight line y=3x+2 with the curve x2+2xy+3y2+4x+8y−11=0 is tan−12√23

Prove that the angle between the straight lines joining the origin to the intersection of the straight line $y = 3 x + 2$ with the curve $x^{2} + 2 x y + 3 y^{2} + 4 x + 8 y - 11 = 0$ is ${tan}^{- 1} \frac{2 \sqrt{2}}{3}$