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A square has one vertex at the vertex of the parabola `y^2=4a 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 `(4a ,4a)` (b) `(4a ,-4a)` `(0,0)` (d) `(8a ,0)`

A

(4a,4a)

B

(4a,-4a)

C

(0,0)

D

(8a,0)

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

AC is one diagonal along the x-axis. Then the other diagonal is BD, where both B and lie on the parabola. Also,

Slope of `AB="tan"(pi)/(4)=1`
If B is `(at^(2),2at)` then
Slope of `AB=(2at)/(at^(2))=(2)/(t)=1`
`:." "t=2`
Therefore, B is (4a,4a) and, hance, D is (4a,-4a).
Clearly, C is (8a,0).
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CENGAGE ENGLISH-PARABOLA-EXERCISE (MULTIPLE CORRECT ANSWER TYPE )
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  14. The line x+ y +2=0 is a tangent to a parabola at point A, intersect t...

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  15. Which of the following line can be normal to parabola y^2=12 x ? x+y-...

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  16. A normal drawn to the parabola =4a x meets the curve again at Q such t...

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  17. A circle is drawn having centre at C (0,2) and passing through focus ...

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  18. From any point P on the parabola y^(2)=4ax, perpebdicular PN is drawn ...

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  19. Let P be the point (1,0) and Q be a point on the locus y^(2)=8x. The l...

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  20. The value(s) of a for which two curves y=ax^(2)+ax+(1)/(24)andx=ay^(2)...

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