Let $P$ be the point $(1, 0)$ and $Q$ be a point on the curve $y^{2} = 8 x$ . Then the locus of the mid-point of $P Q$ is

y^{2} + 4 x + 2 = 0

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y^{2} - 4 x + 2 = 0

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x^{2} - 4 y + 2 = 0

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x^{2} + 4 y + 2 = 0

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Solution

The correct option is B $y^{2} - 4 x + 2 = 0$
Let $R (h, k)$ be the mid-point of $P Q$ .
By mid-point formula,
$Q$ is $(2 h - 1, 2 k)$

Since $Q$ lies on $y^{2} = 8 x$ ,
$(2 k)^{2} = 8 (2 h - 1)$
$\Rightarrow k^{2} = 2 (2 h - 1)$
Hence, locus of $Q$ is
$y^{2} = 2 (2 x - 1)$
$\Rightarrow y^{2} - 4 x + 2 = 0$

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Let P be the point (1,0) and Q be a point on the curve y2=8x. Then the locus of the mid-point of PQ is

The correct option is B y2−4x+2=0Let R(h,k) be the mid-point of PQ. By mid-point formula, Q is (2h−1,2k) Since Q lies on y2=8x, (2k)2=8(2h−1) ⇒k2=2(2h−1) Hence, locus of Q is y2=2(2x−1) ⇒y2−4x+2=0

Let $P$ be the point $(1, 0)$ and $Q$ be a point on the curve $y^{2} = 8 x$ . Then the locus of the mid-point of $P Q$ is