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Question: If the line joining (0,0) to point of intersection of \(5{{x}^{2}}+12xy-6{{y}^{2}}+4x-2y+3=0\) and \...

If the line joining (0,0) to point of intersection of 5x2+12xy−6y2+4x−2y+3=05{{x}^{2}}+12xy-6{{y}^{2}}+4x-2y+3=0 and x+ky−1=0x+ky-1=0 are equally inclined with the coordinate axes then k=?
a)0
b)1
c)-1
d)-2

Explanation

Solution

Hint: In this case we first need to find the point of intersection of 5x2+12xy−6y2+4x−2y+3=05{{x}^{2}}+12xy-6{{y}^{2}}+4x-2y+3=0 and x+ky−1=0x+ky-1=0 in terms of the variable k. Now, as it is given that the line joining the point of intersection to (0,0) is equally inclined with the coordinate axes, we can use the standard equation to determine the value of k.

Complete step-by-step answer:
In this case the angle of the line joining the point of intersection of the curves 5x2+12xy−6y2+4x−2y+3=05{{x}^{2}}+12xy-6{{y}^{2}}+4x-2y+3=0 and x+ky−1=0x+ky-1=0 is given. Therefore, we should first try to find the point of intersection. The point of intersection should satisfy both equations. Therefore, if the point of intersection is (x,y), then it satisfies
x+ky−1=0⇒x+ky1=1...........(1.2)x+ky-1=0\Rightarrow \dfrac{x+ky}{1}=1...........(1.2)
Therefore, to homogenize the equation of the curve i.e. multiply 1 from (1.2) such that each term has the same total power of x and y in the following way
5x2+12xy−6y2+4x(x+ky1)−2y(x+ky1)+3(x+ky1)2=0 ⇒5x2+12xy−6y2+4x2+4kxy−2xy−2ky2+3x2+6kxy+3k2y2=0...............(1.3) \begin{aligned} & 5{{x}^{2}}+12xy-6{{y}^{2}}+4x\left( \dfrac{x+ky}{1} \right)-2y\left( \dfrac{x+ky}{1} \right)+3{{\left( \dfrac{x+ky}{1} \right)}^{2}}=0 \\\ & \Rightarrow 5{{x}^{2}}+12xy-6{{y}^{2}}+4{{x}^{2}}+4kxy-2xy-2k{{y}^{2}}+3{{x}^{2}}+6kxy+3{{k}^{2}}{{y}^{2}}=0...............(1.3) \\\ \end{aligned}
Now, this equation also represents a pair of straight lines passing through the origin. However, from the general equation of a pair of straight lines passing through origin ax2+2hxy+by2=0a{{x}^{2}}+2hxy+b{{y}^{2}}=0 the slopes of the lines are given by
m1=−h+h2−abb and m1=−h−h2−abb{{m}_{1}}=\dfrac{-h+\sqrt{{{h}^{2}}-ab}}{b}\text{ and }{{m}_{1}}=\dfrac{-h-\sqrt{{{h}^{2}}-ab}}{b}
Thus, the line joining the point of intersection of the given curve and the line will be equally inclined with the axes if the coefficient of xy (h in the general equation is 0).
Thus, comparing with equation (1.3), we get
The coefficient of xyxy=12+4k−2+6k=0⇒10k=−10⇒k=−1\text{=}12+4k-2+6k=0\Rightarrow 10k=-10\Rightarrow k=-1
Therefore, the answer to the given question is k=-1 which matches option (c) and thus (c) is the correct option.

Note: We should note that while homogenizing the equation in (1.3) we should be careful to multiply the square of the factor with the constant coefficient term and only the factor with the terms with x and y so that the total power of each term in the resulting equation is the same.