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-A- 9 x2-8 y2-8 y-16- 9 x 2-8 y 2-8 y -16- 8 x 2-9 y 2-9 y -18D- 8 x 2-9 y 2-9 y -18

The correct option is B 9x^2+8y^2 8y=16Given, OA+AP+OP=4 ⇒1+√(x^2+(y 1)^2) + √(x^2+y^2)=4 ⇒(√(x^2+y^2))^2=(3 √(x^2+(y 1)^2))^2 ⇒ x^2+y^2=9+x^2+y^2 2y+1 6√(x^2 ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula

Let O 0,0 and...

Question

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:

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

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

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8 x^{2} - 9 y^{2} + 9 y = 18

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8 x^{2} + 9 y^{2} - 9 y = 18

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Solution

The correct option is B $9 x^{2} + 8 y^{2} - 8 y = 16$
Given,

$O A + A P + O P = 4$
$\Rightarrow 1 + \sqrt{x^{2} + (y - 1)^{2}} + \sqrt{x^{2} + y^{2}} = 4$
$\Rightarrow (\sqrt{x^{2} + y^{2}})^{2} = (3 - \sqrt{x^{2} + (y - 1)^{2}})^{2}$
$\Rightarrow x^{2} + y^{2} = 9 + x^{2} + y^{2} - 2 y + 1 - 6 \sqrt{x^{2} + (y - 1)^{2}}$
$\Rightarrow 3 \sqrt{x^{2} + (y - 1)^{2}} = (5 - y)$
$\Rightarrow 9 (x^{2} + (y - 1)^{2}) = (5 - y)^{2}$
$\Rightarrow 9 x^{2} + 8 y^{2} - 8 y = 16$

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Similar questions

Q. 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

4

, is:

Q. Let

A

and

B

be any two points on the lines represented by

4 x^{2} - 9 y^{2} = 0.

If the area of triangle

O A B

5

(O is origin)

then which of the following is the possible equation of the locus of midpoint of

A B ?

Q. A normal to the hyperbola,

4 x^{2} - 9 y^{2} = 36

meets the co-ordinate axes

x

and

y

A

and

B

, respectively. If the parallelogram

O A B P

(

O

being the origin) is formed, then the locus of

P

is :

Q. PQ is a double ordinate of the parabola

y^{2} = 4 a x

. The locus of the points of trisection of PQ is

Q. If x = 9 is the chord of contact of the hyperbola

x^{2} - y^{2} = 9

, then the equation of the corresponding pair of tangents is

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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:

The correct option is B 9x2+8y2−8y=16Given, OA+AP+OP=4 ⇒1+√x2+(y−1)2 + √x2+y2=4 ⇒(√x2+y2)2=(3−√x2+(y−1)2)2 ⇒x2+y2=9+x2+y2−2y+1−6√x2+(y−1)2 ⇒3√x2+(y−1)2=(5−y) ⇒9(x2+(y−1)2)=(5−y)2 ⇒9x2+8y2−8y=16

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: