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  • Let  be a straight line passing through the origin and<img alt="L_2" src="https://entrancecorner.oncodecogs.com/

Let L_1 be a straight line passing through the origin andL_2be the straight line2x+2y=1. If the intercepts made by the circle3 x^2+2 y^2-x+4 y=0 onL_1 andL_2 are equal, then which of the following equations can representL_1?

 

Option: 1

y=(1 / 2) x \\


Option: 2

y=2 x \\


Option: 3

y=(1 / 4) x \\


Option: 4

y=-4 x


Answers (1)

best_answer

let's consider the general equation of a straight line, which isy=mx+c

Since the line passes through the origin (0, 0), the y-intercept (c) is zero.

Therefore, the equation of L_1can be written asy=mx+c

The intercepts made by the circle equation \begin{aligned} & 3 x^2+2 y^2-x+4 y=0 \\ & \quad 3 x^2+2(m x)^2-x+4(m x)=0 \\ & 3 x^2+2 m^2 x^2-x+4 m x=0 \\ & \left(3+2 m^2\right) x^2+(4 m-1) x=0 \end{aligned}

For the intercepts to be equal, the discriminant of this quadratic equation should be zero.

The discriminant (D) of a quadratic equation a x^2+b x+c=0 is given by,D=b^2-4 a c .

\begin{aligned} & \quad D=(4 m-1)^2-4\left(3+2 m^2\right)(0) \\ & D=0 \end{aligned}

so,

\begin{aligned} & (4 m-1)^2-4\left(3+2 m^2\right)(0)=0 \\ & (4 m-1)^2=0 \end{aligned}

$$ \begin{aligned} & (4 m-1)^2-4\left(3+2 m^2\right)(0)=0 \\ & (4 m-1)^2=0 \\ & 16 m^2-8 m+1=0 \end{aligned} $$ Quadratic equation for $m$ : $$ \begin{aligned} & 16 m^2-8 m+1=0 \\ m= & \left.\left(-b \pm \sqrt{(} b^2-4 a c\right)\right) /(2 a) \\ m & \left.=\left(-(-8) \pm \sqrt{(}(-8)^2-4(16)(1)\right)\right) /(2(16)) \\ m & =(8 \pm \sqrt{(64-64)) / 32} \\ m= & 8 \pm \sqrt{0} / 32 \\ m & =8 / 32 \\ m & =1 / 4 \end{aligned}

Therefore, the slope of the line L_1 is 1/4

Thus, the equation that represents L_1  is y=(1/4)x

Posted by

shivangi.bhatnagar

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