Let $x = a t^{2}$ , $y = 2 a t$ represent the equation of a parabola. Show that the tangents and normals at the points corresponding to $t = 1$ and $t = - 1$ form a squre.

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Solution

The given parabola has equation $y^{2} = 4 a x$ and the given points on the parabola are $A (a, 2 a)$ and $B (a, - 2 a)$ with parameters 1 and -1.
Now, meeting point of tangents $C$ at these points is $(a t_{1} t_{2}, a (t_{1} + t_{2})) = (- a, 0)$
Also, it is known that as the normals are drawn at ends of latus rectum of parabola, they are perpendicular to one another.
Also, the tangent and normal at any point are perpendicular to each other.
Hence, all angles are equal to $90^{\circ}$ . So, the quadrilateral can only be a square or a rectangle.
Now, we just need to prove that any 2 adjacent sides are equal in length.
Hence, $A C = \sqrt{(a + a)^{2} + (2 a)^{2}} = 2 \sqrt{2} a$ and $A B = \sqrt{(a + a)^{2} + (- 2 a)^{2}} = 2 \sqrt{2} a$ .
$∴ A C = A B$
Hence, proved.

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