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Question: If normal at point P on parabola y<sup>2</sup> = 4ax, (a \> 0) meet it again at Q is such a way that...

If normal at point P on parabola y2 = 4ax, (a > 0) meet it again at Q is such a way that OQ is of minimum length where O is vertex of parabola, then DOPQ is

A

A right angled triangle

B

Obtuse angled triangle

C

Acute angled triangle

D

None

Answer

A right angled triangle

Explanation

Solution

\ Normal at P(t1) meets at Q(t2)

t2 = −2t1−t1- \frac{2}{t_{1}} - t_{1}

| t2 | ³ 222\sqrt{2}

For minimum length of OQ, |t2| should be minimum

\ i.e. | t2 | = 222\sqrt{2}

If t2 = −22- 2\sqrt{2} ̃ t1 = 2\sqrt{2}

Slope of OQ = 2t2\frac{2}{t_{2}} = m1 and of OQ = 2t1\frac{2}{t_{1}} = m2

\ m1m2 = –1 \ DOPQ is right angled triangle.