The point P on the parabola $y^{2} = 4 a x$ for which $| P R - P Q |$ is maximum, where R = (-a, 0), Q = (0, a). is

(a, 2a)

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(a, -2a)

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(4a, 4a)

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(4a, -4a)

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Solution

The correct option is A (a, 2a)
We know that any side of the triangle is more than the difference of the remaining two sides so that
$| P R - P Q | \leq R Q$
The required point P will be the point of intersection of the line RQ with parabola which is (a, 2a) as PQ is a tangent to the parabola

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The point P on the parabola y2=4ax for which |PR−PQ| is maximum, where R = (-a, 0), Q = (0, a). is

The correct option is A (a, 2a)We know that any side of the triangle is more than the difference of the remaining two sides so that |PR−PQ|≤RQ The required point P will be the point of intersection of the line RQ with parabola which is (a, 2a) as PQ is a tangent to the parabola

The point P on the parabola $y^{2} = 4 a x$ for which $| P R - P Q |$ is maximum, where R = (-a, 0), Q = (0, a). is

The correct option is A (a, 2a)
We know that any side of the triangle is more than the difference of the remaining two sides so that
$| P R - P Q | \leq R Q$
The required point P will be the point of intersection of the line RQ with parabola which is (a, 2a) as PQ is a tangent to the parabola