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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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Solution

The correct option is A $y^{2} - 4 x + 2 = 0$
$P = (1, 0), Q = (h, k)$

$k^{2} = 8 h$

Let $(α, β)$ be the mid-point of PQ.

$α = \frac{h + 1}{2}, β = \frac{k + 0}{2}$

$2 α - 1 = h, 2 β = k$

$(2 β)^{2} = 8 (2 α - 1)$

$β^{2} = 4 α - 2$

$⟹ y^{2} - 4 x + 2 = 0$

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Similar questions

Q. Let

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and

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be a point on the curve

y^{2} = 8 x

. Then the locus of the mid-point of

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Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Q. Transform to parallel axes through the point

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y^{2} - 4 x + 4 y + 8 = 0

,
and (2)

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

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

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The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is x-y+1=0 is

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

The correct option is A y2−4x+2=0P=(1,0),Q=(h,k)k2=8hLet (α,β) be the mid-point of PQ.α=h+12,β=k+022α−1=h,2β=k(2β)2=8(2α−1)β2=4α−2⟹y2−4x+2=0

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

The correct option is A $y^{2} - 4 x + 2 = 0$
$P = (1, 0), Q = (h, k)$

$k^{2} = 8 h$

Let $(α, β)$ be the mid-point of PQ.

$α = \frac{h + 1}{2}, β = \frac{k + 0}{2}$

$2 α - 1 = h, 2 β = k$

$(2 β)^{2} = 8 (2 α - 1)$

$β^{2} = 4 α - 2$

$⟹ y^{2} - 4 x + 2 = 0$