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Chapter 14: Problem 1215

\(\mathrm{A}(1,0)\) and \(\mathrm{B}(-1,0)\), then the locus of points satisfying \(\mathrm{AQ}-\mathrm{BQ}=\pm 1\) is \(\ldots \ldots\) (a) \(12 \mathrm{x}^{2}+4 \mathrm{y}^{2}=3\) (b) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=3\) (c) \(12 \mathrm{x}^{2}-4 \mathrm{y}^{2}=-3\) (d) \(12 x^{2}+4 y^{2}=-3\)

Short Answer

Expert verified
The locus of points satisfying the given condition is (b) \(12x^2 - 4y^2 = 3\).

Step by step solution

01

Write the distance formula for AQ and BQ

We'll use the distance formula as follows for AQ and BQ: \(AQ = \sqrt{(x - 1)^2 + (y - 0)^2}\) \(BQ = \sqrt{(x + 1)^2 + (y - 0)^2}\)
02

Set up the equation for the distance difference

We are given that the distance difference is either +1 or -1, so we can set up the equation as follows: \(|AQ - BQ| = 1\)
03

Substitute the distance formula expressions in the equation

We'll substitute the expressions we found in Step 1 into the equation from Step 2 and get: \(|\sqrt{(x - 1)^2 + y^2} - \sqrt{(x + 1)^2 + y^2}| = 1\)
04

Eliminate the absolute value

To eliminate the absolute value, we consider two separate cases: Case 1: AQ - BQ = 1 Case 2: AQ - BQ = -1 We will set up the equations for both cases, solve them and see which of the given options corresponds to the solution.
05

Solve for Case 1 (AQ - BQ = 1)

For this case, the equation becomes: \(\sqrt{(x - 1)^2 + y^2} - \sqrt{(x + 1)^2 + y^2} = 1\) Squaring both sides and solving the equation, we get: \(12x^2 - 4y^2 = 3\)
06

Solve for Case 2 (AQ - BQ = -1)

For this case, the equation becomes: \(\sqrt{(x - 1)^2 + y^2} - \sqrt{(x + 1)^2 + y^2} = -1\) Squaring both sides and solving the equation also results in: \(12x^2 - 4y^2 = 3\) Now, we can see both cases lead to the same equation. Therefore, the locus of points satisfying the given condition is: \(12x^2 - 4y^2 = 3\) Comparing this equation with the given options, we find that it matches option (b). So the correct answer is: (b) \(12x^2 - 4y^2 = 3\)

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Most popular questions from this chapter

If two perpendicular lines passing through origin intersect the line \((\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})=1, \mathrm{a} \neq 0, \mathrm{~b} \neq 0\) at \(\mathrm{A}\) and \(\mathrm{B}\), then \(\left\\{1 /(\mathrm{OA})^{2}\right\\}+\left\\{1 /(\mathrm{OB})^{2}\right\\}=\ldots \ldots\) (a) \(\left(1 / \mathrm{a}^{2}\right)-\left(1 / \mathrm{b}^{2}\right)\) (b) \(\left[(\mathrm{ab}) /\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\right]\) (c) \(\left[\left(a^{2}+b^{2}\right) /\left(a^{2} b^{2}\right)\right]\) (d) None of these

The circumcentre of the triangle formed by the lines \(\mathrm{x}+\mathrm{y}=0, \mathrm{x}-\mathrm{y}=0\) and \(\mathrm{x}-7=0\) is \(\ldots \ldots\) (a) \((7,0)\) (b) \((3.5,0)\) (c) \((0,7)\) (d) \((3.5,3.5)\)

The equations of two straight lines which are parallel to \(\mathrm{x}+7 \mathrm{y}+2=0\) and at unit distance from the point \((1,-1)\) are (a) \(x+7 y+6 \pm 4 \sqrt{2}=0\) (b) \(x+7 y+6 \pm 5 \sqrt{2}=0\) (c) \(2 \mathrm{x}+7 \mathrm{y}+6 \pm 5 \sqrt{2}=0\) (d) \(x+y+6 \pm 5 \sqrt{2}=0\)

If the length of perpendicular drawn from \((5,0)\) on \(\mathrm{kx}+4 \mathrm{y}=20\) is 1, then \(\mathrm{k}=\ldots \ldots \ldots\) (a) \(3,(16 / 3)\) (b) \(3,-(16 / 3)\) (c) \(-3,(16 / 3)\) (d) \(-3,-(16 / 3)\)

If the \(\mathrm{y}\) -intercept of the perpendicular bisector of the segment obtained by joining \(\mathrm{P}(1,4)\) and \(\mathrm{Q}(\mathrm{k}, 3)\) is \(-4\) then \(\mathrm{k}=\ldots \ldots\) (a) 1 (b) 2 (c) \(-2\) (d) \(-4\)
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