SATHEE: Circle 1 Question 2

Circle 1 Question 2

2. Let $O (0, 0)$ and $A (0, 1)$ be two fixed points, then the locus of a point $P$ such that the perimeter of $△ A O P$ is 4 , is

(2019 Main, 8 April I)

(a) $8 x^{2} - 9 y^{2} + 9 y = 18$

(b) $9 x^{2} - 8 y^{2} + 8 y = 16$

(d) $8 x^{2} + 9 y^{2} - 9 y = 18$

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Answer:

Correct Answer: 2. (c)

Solution:

Given vertices of $△ A O P$ are $O (0, 0)$ and $A (0, 1)$

Let the coordinates of point $P$ are $(x, y)$ .

Clearly, perimeter $= O A + A P + O P = 4$ (given)

$\Rightarrow \sqrt{(0 - 0)^{2} + (0 - 1)^{2}} + \sqrt{(0 - x)^{2} + (1 - y)^{2}} + \sqrt{x^{2} + y^{2}} = 4$

$\Rightarrow 1 + \sqrt{x^{2} + (y - 1)^{2}} + \sqrt{x^{2} + y^{2}} = 4$

$\Rightarrow \sqrt{x^{2} + y^{2} - 2 y + 1} + \sqrt{x^{2} + y^{2}} = 3$

$\Rightarrow \sqrt{x^{2} + y^{2} - 2 y + 1} = 3 - \sqrt{x^{2} + y^{2}}$

$\Rightarrow x^{2} + y^{2} - 2 y + 1 = 9 + x^{2} + y^{2} - 6 \sqrt{x^{2} + y^{2}}$

[squaring both sides]

$\Rightarrow 1 - 2 y = 9 - 6 \sqrt{x^{2} + y^{2}}$

$\Rightarrow 6 \sqrt{x^{2} + y^{2}} = 2 y + 8$

$\Rightarrow 3 \sqrt{x^{2} + y^{2}} = y + 4$

$\Rightarrow 9 (x^{2} + y^{2}) = (y + 4)^{2}$

$\Rightarrow 9 x^{2} + 9 y^{2} = y^{2} + 8 y + 16$

$\Rightarrow 9 x^{2} + 8 y^{2} - 8 y = 16$

Thus, the locus of point $P (x, y)$ is

$9 x^{2} + 8 y^{2} - 8 y = 16$

Circle 1 Question 2

2. Let $O (0, 0)$ and $A (0, 1)$ be two fixed points, then the locus of a point $P$ such that the perimeter of $△ A O P$ is 4 , is

Answer:

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Circle 1 Question 2

2. Let O(0,0) and A(0,1) be two fixed points, then the locus of a point P such that the perimeter of △AOP is 4 , is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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2. Let $O (0, 0)$ and $A (0, 1)$ be two fixed points, then the locus of a point $P$ such that the perimeter of $△ A O P$ is 4 , is