Question

The angle between the lines joining the origin to the points of intersection of the line y = 3x + 2 with the curve $x^2$ + 2xy + ${3y}^2$ + 4x + 8y = 11, is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The equation of the given straight line is
$Y=3x+2\Rightarrow \frac{y-3x}{2}=1......\left(i\right)$
The equation of the given curve is
$x^2+2xy+3y^2+4x+8y-11=0........\left(ii\right)$
The combined equation of the straight lines joining the origin to the points of intersection of Eqs. (i) and (ii) is a homogeneous equation of second degree obtained with the help of Eqs. (i) and (ii). Making the Eq. (ii) homogeneous of the second degree in x and y with the help of Eq. (i), we get
$x^2+2xy+3y^2+4x\left(\frac{y-3x}{2}\right)+8y\left(\frac{y-3x}{2}\right)-11{\left(\frac{y-3x}{2}\right)}^2=0\Rightarrow 4x^2+8xy+12y^2+2(8y^2-12x^2-20xy)-11(y^2-6xy+9x^2)=0\Rightarrow 7x^2-2xy-y^2=0$
This is the required equation. On comparing the equation with $a{x}^2+2hxy+b{y}^2=0$, we obtain
a = 7, b = −1 and h = −1.
Let $\theta$ be the required angle. Then,
$tan\theta =\frac{2\sqrt{h^2-ab}}{a+b}\Rightarrow tan\theta =\frac{2\sqrt{1+7}}{a+b}\Rightarrow \theta =ta{n}^{-}\left(\frac{2\sqrt{2}}{3}\right)$
Hence, (c) is the correct answer.