Q52 / 160 · AP EAMCET 2012 · Mathematics

If the line $2 x+5 y=12$ intersects the ellipse $4 x^2+5 y^2=20$ in two distinct points $A$ and $B$, then mid-point of $A B$ is

  • A
    $(0,1)$
  • B
    $(1,2)$
  • C
    $(1,0)$
  • Correct answer
    D
    None of these

Solution

Since, the line $2 x+5 y=12$ intersect the ellipse $4 x^2+5 y^2=20$ $\begin{array}{ll}\therefore & 4\left(\frac{12-5 y}{2}\right)^2+5 y^2=20 \\ \Rightarrow & 144+25 y^2-120 y+5 y^2=20 \\ \Rightarrow & 30 y^2-120 y+124=0 \\ \Rightarrow & 15 y^2-60 y+62=0 \\ \therefore & y=\frac{60 \pm \sqrt{(60)^2-4 \times 15 \times 62}}{2(15)} \\ & =\frac{60 \pm \sqrt{3600-3720}}{30}=\frac{60 \pm \sqrt{-120}}{30}\end{array}$ Hence, no real value of $y$ exist. Hence, no intersection is possible.

Hence, the correct answer is option: 'D'.

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