AP EAMCET · Maths · Straight Lines
If the line \(x+2 y=k\) intersects the curve \(x^2-x y+y^2+3 x+3 y-2=0\) at two points \(A\) and \(B\) and if \(O\) is the origin, then the condition for \(\angle \mathrm{AOB}=90^{\circ}\) is
- A \(k^2+k+1=0\)
- B \(k^2-2 k+10=0\)
- C \(2 k^2+9 k-10=0\)
- D \(3 k^2+8 k-1=0\)
Answer & Solution
Correct Answer
(C) \(2 k^2+9 k-10=0\)
Step-by-step Solution
Detailed explanation
We have, \[ x^2-x y+y^2+3 x+3 y-2=0 \] and \[ x+2 y=k \Rightarrow \frac{x+2 y}{k}=1 \] By homogeneous of Eq. (i), we get \[ \begin{aligned} x^2-x y+y^2+3 x & \left(\frac{x+2 y}{k}\right) \\ & +3 y\left(\frac{x+2 y}{k}\right)-2\left(\frac{x+2 y}{k}\right)^2=0 \end{aligned} \]…
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