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 [2018]

Question

Let $S$ be the circle in the $xy$ -plane defined by the equation $x^{2} + y^{2} = 4$ . [2018]

Q. 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

Smart Achievers · Accepted Answer

(4)

$Let point P be (2 \cos θ, 2 \sin θ)$

$Tangent at P is$ $x \cos θ + y \sin θ = 2$

$∴$ $M (\frac{2}{\cos θ}, 0) and N (0, \frac{2}{\sin θ})$

$∴$ $Midpoint of M N =$ $(\frac{1}{\cos θ}, \frac{1}{\sin θ})$

$For locus of midpoint (x, y) of M N,$

$x = \frac{1}{\cos θ}, y = \frac{1}{\sin θ}$

$\Rightarrow \frac{1}{x^{2}} + \frac{1}{y^{2}} = 1 \Rightarrow x^{2} + y^{2} = x^{2} y^{2}$

Solution

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

2 $x^{2 / 3} + y^{2 / 3} = 2^{4 / 3}$

3 $x^{2} + y^{2} = 2 x y$

4 $x^{2} + y^{2} = x^{2} y^{2}$

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Solution

Q. Let S be the circle in the xy-plane defined by the equation x2+y2=4. [2018] Q. 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

1 (x+y)2=3xy

2 x2/3+y2/3=24/3

3 x2+y2=2xy

4 x2+y2=x2y2

Ans. (4) Let point P be (2cosθ,2sinθ) Tangent at P is xcosθ+ysinθ=2 ∴ M(2cosθ,0) and N(0,2sinθ) ∴ Midpoint of MN=(1cosθ,1sinθ) For locus of midpoint (x,y) of MN, x=1cosθ, y=1sinθ ⇒1x2+1y2=1⇒x2+y2=x2y2

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

2 $x^{2 / 3} + y^{2 / 3} = 2^{4 / 3}$

3 $x^{2} + y^{2} = 2 x y$

4 $x^{2} + y^{2} = x^{2} y^{2}$