Question

The value of $\lambda$ such that the line joining the origin to the points of the intersection of the line $x+y=1$ and the curve $x^2+y^2+x-2y-\lambda =0$ are mutually perpendicular.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for correct option

Step 1: Finding the slopes of line and curve
Given curve, $x^2+y^2+x-2y-\lambda =0$ on the line $x+y=1$

We can convert the given curve into an homogeneous equation as follows,

$$\begin{gathered} \Rightarrow x^2+y^2+(x-2y)\times 1-\lambda \times {\left(1\right)}^2=0 \\ \Rightarrow x^2+y^2+\left(x-2y\right)\left(x+y\right)-\lambda {(x+y)}^2=0 \\ \Rightarrow x^2+y^2+x^2+xy-2xy-2y^2-\lambda (x^2+y^2+2xy)=0 \\ \Rightarrow x^2+y^2+x^2+xy-2xy-2y^2-\lambda x^2-\lambda y^2-2\lambda xy=0 \\ \Rightarrow x^2(2-\lambda )+y^2(-1-\lambda )+2xy(-1-\lambda )=0 \end{gathered}$$

Since, the line joining the origin to the points of intersection are perpendicular, so $m_1\times m_2=-1$

where, $m_1$ is the slope of $x^2+y^2+x-2y-\lambda =0$, that is, $-\frac{2-\lambda }{1+\lambda }$ [Slope of a line $=\frac{y}{x}$ (coordinates)]

and similarly, $m_2$ is the slope of $x+y=1$, that is, $1$.

Step 2: Finding the value of $\lambda$

Solving the equation of slope, we get:

$$\begin{gathered} \frac{2-\lambda }{1+\lambda }=1 \\ \Rightarrow 2-\lambda =1+\lambda \\ \Rightarrow 2-\lambda -1-\lambda =0 \\ \Rightarrow 1-2\lambda =0 \\ \Rightarrow \lambda =\frac{1}{2} \end{gathered}$$

Hence, the correct answer is option A.