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Chapter 15: Problem 1395

Let \(\mathrm{P}\) be the point \((1,0)\) and \(\mathrm{Q}\) a point on the locus \(\mathrm{y}^{2}=8 \mathrm{x}\). The locus of mid-point of \(P Q\) is (a) \(\mathrm{y}^{2}+4 \mathrm{x}+2=0\) (b) \(y^{2}-4 x+2=0\) (c) \(x^{2}-4 y+2=0\) (d) \(x^{2}+4 y+2=0\)

Short Answer

Expert verified
The locus of the midpoint M is \(y^2 - 4x + 2 = 0\). The correct option is choice (b).

Step by step solution

01

Identify the coordinates of point Q

Given the equation of locus \(y^2 = 8x\), let point Q have coordinates \((x_Q, y_Q)\) such that \(y_Q^2 = 8x_Q\).
02

Calculate the midpoint of segment PQ

Let M be the midpoint of segment PQ, and has coordinates \((x_M, y_M)\). To find the coordinates \(x_M\) and \(y_M\), we can use the midpoint formula: \[x_M = \frac{x_P + x_Q}{2}\] \[y_M = \frac{y_P + y_Q}{2}\] Substitute \(x_P = 1\) and \(y_P = 0\) to find the coordinates of M in terms of coordinates of Point Q: \[x_M = \frac{1 + x_Q}{2}\] \[y_M = \frac{0 + y_Q}{2} = \frac{y_Q}{2}\]
03

Derive the equation of the locus of midpoint M

Replace \(x_Q\) and \(y_Q\) from \(y_Q^2 = 8x_Q\) with the equivalent expressions in terms of \(x_M\) and \(y_M\): Substitute \(x_Q = 2x_M - 1\): \[y_Q^2 = 8(2x_M - 1)\] Substitute \(y_Q = 2y_M\): \[(2y_M)^2 = 16x_M - 8\] Simplify and rewrite the equation: \[4y_M^2 = 16x_M - 8\] \[y_M^2 - 4x_M + 2 = 0\] The locus of the midpoint M is \(y^2 - 4x + 2 = 0\). The correct option is choice (b).

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Most popular questions from this chapter

Two tangents to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=4\) at the points \(\mathrm{A}\) and \(\mathrm{B}\) meet at \(\mathrm{P}(-4,0)\). The area of the quadrilateral \(\mathrm{PAOB}\), where \(\mathrm{O}\) is the origin is (a) \(4 \sqrt{3}\) (b) 4 (c) \(6 \sqrt{2}\) (d) \(2 \sqrt{3}\)

The shortest distance between the line \(\mathrm{x}-\mathrm{y}+1=0\) and the curve \(\mathrm{x}=\mathrm{y}^{2}\) is \(\ldots \ldots \ldots\) (a) \([(3 \sqrt{2}) / 5]\) (b) \([(2 \sqrt{3}) / 8]\) (c) \([(3 \sqrt{2}) / 8]\) (d) \([(2 \sqrt{2}) / 5]\)

If a chord of the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\), passing through its focus \(\mathrm{F}\) meets it in \(\mathrm{P}\) and \(\mathrm{Q}\), then \([1 /(\mid \mathrm{FP})]+[1 /(\mathrm{FQ})]=\ldots \ldots \ldots\) (a) \((1 / a)\) (b) (2/a) (c) \((4 / \mathrm{a})\) (d) \((1 / 2 \mathrm{a})\)

The equations to the common tangents to the two hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) and are \(\left(y^{2} / a^{2}\right)-\left(x^{2} / b^{2}\right)=1\) (a) \(\mathrm{y}=\pm \mathrm{x} \pm \sqrt{\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}\) (b) \(\mathrm{y}=\pm \mathrm{x} \pm \sqrt{\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}\) (c) \(\mathrm{y}=\pm \mathrm{x} \pm \sqrt{\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}\) (d) \(\mathrm{y}=\pm \mathrm{x} \pm\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\)

The eccentricity of an ellipse, with its centre at the origin, is \((1 / 2)\). If one of the directories is \(\mathrm{x}=4\), then equation of an ellipse is (a) \(3 \mathrm{x}^{2}+4 \mathrm{y}^{2}=1\) (b) \(3 x^{2}+4 y^{2}=12\) (c) \(4 x^{2}+3 y^{2}=12\) (d) \(4 x^{2}+3 y^{2}=1\)
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