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QJEEAdv18Ma5338: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma5338 Solution: Let S be the circle in the XY-plane defined the equation x 2 +…

Let S be the circle in the XY-plane defined the equation x2 + y2 = 4.

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

Multiple Choice Options

  • A (x + y)2 = 3xy
  • B x2/3 + y2/3 = 24/3
  • C x2 + y2 = 2xy
  • D x2 + y2 = x2y2

Step-by-Step Verified Solution

We have,

x2 + y2 = 4

Let P(2 cos θ\theta , 2 sin θ\theta ) be a point on a circle.

∴\therefore Tangent at P is

2 cos θ\theta x + 2 sin θ\theta y = 4

⇒\Rightarrow x cos θ\theta + y sin θ\theta = 2

JEE Advanced 2018 Paper 1 Offline Mathematics - Ellipse Question 7 English Explanation
∴\therefore The coordinates at M(2cos⁡θ,0)M\left( {{2 \over {\cos \theta }},0} \right) and N(0,2sin⁡θ)N\left( {0,{2 \over {\sin \theta }}} \right)

Let (h, k) is mid-point of MN

∴\therefore h=1cos⁡θh = {1 \over {\cos \theta }} and k=1sin⁡θk = {1 \over {\sin \theta }}

⇒\Rightarrow cos⁡θ=1h\cos \theta = {1 \over h}

and sin⁡θ=1k\sin \theta = {1 \over k}

⇒\Rightarrow cos⁡2θ+sin⁡2θ=1h2+1k2{\cos ^2}\theta + {\sin ^2}\theta = {1 \over {{h^2}}} + {1 \over {{k^2}}}

⇒1=h2+k2h2.k2\Rightarrow 1 = {{{h^2} + {k^2}} \over {{h^2}.{k^2}}}

h2+k2=h2k2{h^2} + {k^2} = {h^2}{k^2}

∴\therefore Mid-point of MN lie on the curve

x2+y2=x2y2{x^2} + {y^2} = {x^2}{y^2}

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