AP EAMCET · Maths · Area Under Curves
Area of the region enclosed by the curves \(3 x^2-y^2-2 x y\) \(+4 x+1=0\) and \(3 x^2-y^2-2 x y+6 x+2 y=0\) is
- A \(\frac{3}{4}\)
- B \(\frac{1}{4}\)
- C 1
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Given equation of curve is \(\begin{aligned} & 3 x^2+(4-2 y) x+1-y^2=0 \\ & \therefore x=\frac{(2 y-4) \pm \sqrt{(4-2 y)^2-4 \times 3\left(1-y^2\right)}}{6} \\ & -\frac{2(y-2) \pm \sqrt{4(2 y-1)^2}}{6}=\frac{(y-2) \pm(2 y-1)}{3} \end{aligned}\) \(\therefore\) Equation of pair of…
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- Match the items of List - I with those of the entires of List - II
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