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Maths-

General

Easy

Question

Let y² = 4ax be parabola and PQ be a focal chord of parabola. Let T be the point of intersection of tangents at P and Q. Then

area of circumcircle of ΔPQT is $(\frac{π (P Q)^{2}}{4})$
orthocenter of ΔPQT will lie on tangent at vertex.
incenter of ΔPQT will be vertex of parabola.
incentre of ΔPQT will lie on directrix of parabola.

The correct answer is: area of circumcircle of ΔPQT is $open parentheses fraction numerator pi left parenthesis P Q right parenthesis to the power of 2 end exponent over denominator 4 end fraction close parentheses$

for focal chord t₁t₂ = – 1
Tangent drawn at the extremities of focal chord are perpendicular and meet at directrix
$straight angle PTQ$ = 90º
Hence PQ is diameter of circum circle of ΔPTQ.
2r = PQ $rightwards double arrow$ r = (PQ/2)
Area of circum circle $pi r squared$ = $fraction numerator pi left parenthesis P Q right parenthesis to the power of 2 end exponent over denominator 4 end fraction$

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