Question

Let O be the origin and A be a point on the curve y2=4x. Then the locus of the mid point of OA is :
(a)x2=4y(c)y2=16x(b)x2=2y(d)y2=2x

Answer 1

Answer:

ANSWER IN ATTACHED PHOTO

Answer 2

The locus of the mid point of OA is y² = 2x

Given :

Let O be the origin and A be a point on the curve y² = 4x

To find :

The locus of the mid point of OA is :

(a) x² = 4y

(b) x² = 2y

(c) y² = 16x

(d) y² = 2x

Solution :

Step 1 of 3 :

Write down the given equation of the curve

Here the given equation of the curve is

y² = 4x

Now O be the origin and A be a point on the curve

Let the coordinate of the point A is (m, n)

Since A(m, n) is a point on the curve y² = 4x

∴ n² = 4m - - - - - - (1)

Step 2 of 3 :

Find the locus of the mid point of OA

Let P(h, k) be the mid point of OA

Then we get

$\displaystyle \sf{(h ,k) = \bigg( \frac{0 + m}{2} \: , \: \frac{0 + n}{2} \bigg)}$

Which gives

$\displaystyle \sf{ h = \frac{0 + m}{2} \: , \: k = \frac{0 + n}{2} }$

$\displaystyle \sf{ \implies h = \frac{m}{2} \: , \: k = \frac{ n}{2} }$

$\displaystyle \sf{ \implies m = 2h \: , \: n = 2k}$

Putting the value of m and n in Equation 1 we get

$\displaystyle \sf{ {(2k)}^{2} = 4 \times (2h)}$

$\displaystyle \sf{ \implies 4 {k}^{2} = 8h}$

$\displaystyle \sf{ \implies {k}^{2} = 2h}$

So the locus of the mid point of OA is y² = 2x

Step 3 of 3 :

Choose the correct option

Checking for option (a) x² = 4y

Since the locus of the mid point of OA is y² = 2x

So option (a) is not correct

Checking for option (b) x² = 2y

Since the locus of the mid point of OA is y² = 2x

So option (b) is not correct

Checking for option (c) y² = 16x

Since the locus of the mid point of OA is y² = 2x

So option (c) is not correct

Checking for option (d) y² = 2x

Since the locus of the mid point of OA is y² = 2x

So option (d) is correct

Conclusion : Hence the correct option is (d) y² = 2x

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