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Need explanation for: - Co-ordinate geometry - JEE Main-3

Let $O(0,0)$ and $A(0,1)$ be two fixed points. Then the locus of a point P such that the perimeter of $\Delta AOP$ is 4, is :

• Option 1)

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

• Option 2)

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

• Option 3)

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

• Option 4)

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

Views

$0(0,0)\; \; A(0,1)\; \; \; \; Let \; P = (x,y)$

then perimeter = $OA+AP+PO$

$=OA=1$

$\\AP=\sqrt{x^{2}+(y-1)^{2}}\\\; \; \; \; \\PO=\sqrt{x^{2}+y^{2}}$

$=OA+AP+PO=4\; \; \; \; \; \; (given)$

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

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

$\left ( \sqrt{x^{2}+(y-1)^{2}} \right )^{2}=\left ( 3-\sqrt{x^{2}+y^{2}} \right )^{2}$

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

$-2y-8=-6\sqrt{(x^{2}+y^{2})}$

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

Squaring both sides $=y^{2}+8y+16=9(x^{2}+y^{2})$

$=9x^{2}+8y^{2}-8y-16=0$

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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