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Question

If the tangents and normals at the extremities of a focal chord of any standard parabola intersect at S(x1,y1) and R(x2,y2) respectively, then

A
If the equation of parabola is y2=±4ax, then y1=y2
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B
If the equation of parabola is x2=±4ay, then x1=x2
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C
the circle with focal chord as diameter will always pass through the points R and S
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D
the circle with RS as diameter will always pass through extemities of focal chord
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Solution

The correct option is D the circle with RS as diameter will always pass through extemities of focal chord
Let P(at21,2at1) and Q(at22,2at2) be the extremities of a focal chord of the parabola y2=4ax. The tangents at P and Q intersect at S(at1t2,a(t1+t2))S(a,a(t1+t2)).
[PQ is a focal chord]
x1=a and y1=a(t1+t2)

The normals at P and Q intersect at
R(2a+a(t21+t22+t1t2),at1t2(t1+t2))R(a+a(t21+t22),a(t1+t2))
x2=a+a(t21+t22) and y2=a(t1+t2)
Clearly, y1=y2
Similarly it can be proved for the parabola y2=4ax
and in same way if we consider the parabola of the form x2=±4ay, then we get x1=x2

Now from above figure in the quadrilateral PRQS
P+Q=S+R=180
PRQS will be cyclic quadrilateral

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