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Chapter 14: Problem 1280

The locus of the variable point whose distance from \((-2,0)\) is \((2 / 3)\) times its distance from the line \(\mathrm{x}=-(9 / 2)\) is (a) ellipse (b) parabola (c) circle (d) hyperbola

Short Answer

Expert verified
The locus of the variable point is an ellipse, which can be derived from the equation \[ \frac{5}{9}x^2 - \frac{4}{3}x + y^2 - 1 = 0 \].

Step by step solution

01

Distance formula

Let the point be \((x, y)\). We need to find the distance between this point and \((-2,0)\), and then between the point and the line \(x=-(9 / 2)\) using the distance formula. The distance between \((x,y)\) and \((-2,0)\) is given by the formula: \[ D_1 = \sqrt{(x + 2)^2 + (y - 0)^2} \] The distance between \((x,y)\) and the line \(x=-(9 / 2)\) is given by the formula: \[ D_2 = |x - (-\frac{9}{2})| = |x + \frac{9}{2}|\]
02

Equate the ratios and simplify

Now we are given that the distance from the point to \((-2,0)\) is \((2 / 3)\) times its distance from the line \(x=-(9 / 2)\). Thus, \[ D_1 = \frac{2}{3}D_2 \] Substitute the values of \(D_1\) and \(D_2\) from step 1. \[ \sqrt{(x + 2)^2 + y^2} = \frac{2}{3}|x + \frac{9}{2}|\] Square both sides to eliminate the square root: \[ (x + 2)^2 + y^2 = \frac{4}{9}(x + \frac{9}{2})^2 \]
03

Expand and simplify

Expand both sides of the equation and simplify to determine the locus. \[ x^2 + 4x + 4 + y^2 = \frac{4}{9}(x^2 + 9x + \frac{81}{4}) \] Multiply both sides of the equation by 9: \[ 9x^2 + 36x + 36 + 9y^2 = 4(x^2 + 9x + \frac{81}{4}) \] Combining terms and simplifying, \[ 5x^2 - 12x + 9y^2 - 9 = 0 \] Divide both sides of the equation by 9: \[ \frac{5}{9}x^2 - \frac{4}{3}x + y^2 - 1 = 0 \] By looking at the equation, we can see that the locus is an ellipse because it fits the form, \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Thus, the answer is (a) ellipse.

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Most popular questions from this chapter

The in centre of a triangle whose vertices \(\mathrm{A}(2,4), \mathrm{B}(2,6)\) and \(\mathrm{C}(2+\sqrt{3}, 5)\) is.... (a) \([2+(1 / \sqrt{3}), 5]\) (b) \([1+\\{1 /(2 \sqrt{3})\\},(5 / 2)]\) (c) \((2,5)\) (d) None of these

For the line \(\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right), \mathrm{m}\) and \(\mathrm{x}_{1}\) are fixed values, if different lines are drawn according to the different value of \(\mathrm{y}_{1}\), then all such lines would be \(\ldots\) (a) all lines intersect the line \(\mathrm{x}=\mathrm{x}_{1}\) (b) all lines pass through one fixed point (c) all lines are parallel to the line \(\mathrm{y}=\mathrm{x}_{1}\) (d) all lines will be the set of perpendicular lines

A square of side a lies above the \(\mathrm{x}\) -axis and has one vertex at the origin. The side passing through the origin makes an angle a \(\alpha[0<\alpha<(\pi / 4)]\) with the positive direction of \(\mathrm{x}\) -axis. The eq. of its diagonal not passing through the origin is: (A) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\sin \alpha-\cos \alpha)=\mathrm{a}\) (B) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\sin \alpha+\cos \alpha)=\mathrm{a}\) (C) \(\mathrm{y}(\cos \alpha+\sin \alpha)+\mathrm{x}(\cos \alpha-\sin \alpha)=\mathrm{a}\) (D) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\)

Locus of the centroid of the triangle whose vertices are (acost, asint), (bsint, - bcost) and \((1,0)\), where \(t\) is a parameter is (a) \((3 x-1)^{2}+(3 y)^{2}=a^{2}-b^{2}\) (b) \((3 x-1)^{2}+(3 y)^{2}=a^{2}+b^{2}\) (c) \((3 x+1)^{2}+(3 y)^{2}=a^{2}+b^{2}\) (d) \((3 x+1)^{2}+(3 y)^{2}=a^{2}-b^{2}\)

Area of a quadrilateral formed by the lines \(|\mathrm{x}|+|\mathrm{y}|=2\) is (a) 8 (b) 6 (c) 3 (d) None
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