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NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

The locus of the mid-point of all chords of the parabola $y^2 = 4x$ , which are drawn through its vertex is:

Choose the correct answer:

(Correct Answer)

C.
$x^2 + 4y^2 = 16$

D.
$x^2 = 2y$

Explanation

1. Identify the Points

The given parabola is $y^2 = 4x$ .

The vertex of this parabola is $V(0, 0)$ .

Let $P(x_1, y_1)$ be any point on the parabola. Since $P$ lies on $y^2 = 4x$ , it satisfies:

$y_1^2 = 4x_1$

2. Define the Mid-point

Let $M(h, k)$ be the mid-point of the chord $VP$ , where $V$ is the vertex $(0,0)$ and $P$ is $(x_1, y_1)$ .

Using the mid-point formula:

$h = \frac{0 + x_1}{2} \implies x_1 = 2h$

$k = \frac{0 + y_1}{2} \implies y_1 = 2k$

3. Substitute into the Parabola Equation

Now, substitute the values of $x_1$ and $y_1$ into the equation of the parabola ( $y_1^2 = 4x_1$ ):

$(2k)^2 = 4(2h)$

$4k^2 = 8h$

4. Simplify and Find the Locus

Divide both sides by $4$ :

$k^2 = 2h$

To find the final equation of the locus, replace $(h, k)$ with $(x, y)$ :

$y^2 = 2x$

Conclusion

The locus of the mid-points of the chords drawn through the vertex of the parabola $y^2 = 4x$ is $y^2 = 2x$ .

Correct Option: (b)

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