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In the adjacent figure a parabola is drawn to pass through the vertices $\mathrm{B}, \mathrm{C}\: \text{and} \: \mathrm{D}$ of the square $\mathrm{ABCD}$ . If $\mathrm{\mathrm{A}(2,1), \mathrm{C}(2,3)}$ then focus of this parabola is -

Option: 1
$\left(1, \frac{11}{4}\right)$

Option: 2
$\left(2, \frac{11}{4}\right)$

Option: 3
$\left(3, \frac{13}{4}\right)$

Option: 4
$\left(2, \frac{13}{4}\right)$

Answers (1)

best_answer

Clearly $\mathrm{AC}$ is parallel to y-axis. It's midpoint is $(2,2).$ Thus $\mathrm{B} \equiv(1,2).$

Parabola will be in the form of $\mathrm{(\mathrm{x}-2)^2=\lambda(\mathrm{y}-3).}$

It passes through $(3,2)$

$\mathrm{\Rightarrow 1=-\lambda}$ . Thus parabola is $\mathrm{(\mathrm{x}-2)^2=-1(\mathrm{y}-3).}$

It focus is $\mathrm{x-2=0 . y-3=-\frac{1}{4}}$ , i.e., $\mathrm{\left(2, \frac{11}{4}\right).}$

Hence option 2 is correct.

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manish

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