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Question

A square has one vertex at the vertex of the parabola $$ y^{2}=4 a x $$ and the diagonal through the vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinates of the vertices of the square are


A
(4a,4a)
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B
(4a,4a)
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C
(0,0)
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D
(8a,0)
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Solution

The correct options are
A (0,0)
B $$ (4 a, 4 a) $$
C $$ (4 a,-4 a) $$
D $$ (8 a, 0) $$
$$ A C $$ is one diagonal along $$ x $$ -axis, then the other diagonal is $$ B D $$ where both $$ B $$ and $$ D $$ lie on parabola.
 Also slope of $$ A B $$ is $$ \tan \dfrac{\pi}{4}=1 . $$
 If $$ B $$ is $$ \left(a t^{2}, 2 a t\right) $$ then the slope of $$ A B $$
$$ \therefore =\dfrac{2 a t}{a t^{2}}=\dfrac{2}{t}=1 $$
$$\therefore t=2$$
Therefore, $$ B $$ is $$ (4 a, 4 a) $$ and hence $$ D $$ is $$ (4 a,-4 a) $$
Clearly, $$ C $$ is $$ (8 a, 0) $$.

1634151_1768137_ans_bac563bca0f041c9816424c536f04de9.png

Mathematics

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