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QJEEM19Maf800: JEE Main 2019 Mathematics - Class 11 JEEM19Maf800 Solution: Let O(0, 0) and A(0, 1) be two fixed points. Then the locus of…

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

Multiple Choice Options

  • A 9x2 + 8y2 – 8y = 16
  • B 8x2 – 9y2 + 9y = 18
  • C 8x2 + 9y2 – 9y = 18
  • D 9x2 – 8y2 + 8y = 16

Step-by-Step Verified Solution

JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 103 English Explanation
Let point C(h, k)

Given, AB + BC + AC = 4

From graph, AB = 1

∴\therefore BC + AC = 3

⇒\Rightarrow h2+k2+h2+(k−1)2=3\sqrt {{h^2} + {k^2}} + \sqrt {{h^2} + {{\left( {k - 1} \right)}^2}} = 3

⇒\Rightarrow h2+(k−1)2{{h^2} + {{\left( {k - 1} \right)}^2}} = 9 + h2+k2{{h^2} + {k^2}} - 6h2+k26\sqrt {{h^2} + {k^2}}

⇒\Rightarrow 6h2+k2=2k+86\sqrt {{h^2} + {k^2}} = 2k + 8

⇒\Rightarrow 9(h2+k2)=k2+8k+169\left( {{h^2} + {k^2}} \right) = {k^2} + 8k + 16

⇒\Rightarrow 9h2+8k2−8k−16=09{h^2} + 8{k^2} - 8k - 16 = 0

∴\therefore Locus of point C will be,

9x2 + 8y2 – 8y = 16

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