Let P and Q be distinct points on the parabola such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle is , then which of the following is (are) the coordinates of

Question

Let P and Q be distinct points on the parabola $y^{2} = 2 x$ such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle $∆ O P Q$ is $3 \sqrt{2}$ , then which of the following is (are) the coordinates of P? [2015]

Smart Achievers · Accepted Answer

(1, 4)

Let point P be in first quadrant and lying on parabola $y^{2} = 2 x$ be $(\frac{a^{2}}{2}, a) .$ Let Q be the point $(\frac{b^{2}}{2}, b) .$ Clearly $a > 0 .$

$∵$ $P Q is the diameter of circle through P, O and Q$

$∴$ $∠ P O Q = 90 ° \Rightarrow \frac{a}{a^{2} / 2} \times \frac{b}{b^{2} / 2} = - 1 \Rightarrow a b = - 4$

$\Rightarrow b is negative (∵ a > 0)$ Given area $(∆ P O Q) = 3 \sqrt{2}$

$\Rightarrow \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ \frac{a^{2}}{2} & a & 1 \\ \frac{b^{2}}{2} & b & 1 \end{matrix} |$ $= 3 \sqrt{2}$

$\Rightarrow \frac{1}{4} a b (a - b) = \pm 3 \sqrt{2} \Rightarrow a - b = \pm 3 \sqrt{2}$

As $a$ is positive and $b$ is negative, we have $a - b = 3 \sqrt{2}$

$\Rightarrow a + \frac{4}{a} = 3 \sqrt{2} (∵ a b = - 4)$

$\Rightarrow a^{2} - 3 \sqrt{2} a + 4 = 0 \Rightarrow (a - 2 \sqrt{2}) (a - \sqrt{2}) = 0$

$∴$ $a = 2 \sqrt{2}, \sqrt{2}$

$∴$ $Point P can be (\frac{{(2 \sqrt{2})}^{2}}{2}, 2 \sqrt{2}) or (\frac{{(\sqrt{2})}^{2}}{2}, \sqrt{2})$

$i.e. (4, 2 \sqrt{2}) or (1, \sqrt{2})$

Solution

1 $(4, 2 \sqrt{2})$

2 $(9, 3 \sqrt{2})$

3 $(\frac{1}{4}, \frac{1}{\sqrt{2}})$

4 $(1, \sqrt{2})$

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Solution

Q. Let P and Q be distinct points on the parabola y2=2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ∆OPQ is 32, then which of the following is (are) the coordinates of P [2015]

1 (4,22)

2 (9,32)

3 (14,12)

4 (1,2)

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1 $(4, 2 \sqrt{2})$

2 $(9, 3 \sqrt{2})$

3 $(\frac{1}{4}, \frac{1}{\sqrt{2}})$

4 $(1, \sqrt{2})$