Let $S$ be the circle in the $xy$-plane defined by the equation $x^{2}+y^{2}=4$. 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 $M N$ must lie on the curve

(D) $x^{2}+y^{2}=x^{2} y^{2}$

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

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Circle

Passage

Let

S

be the circle in the

x y

-plane defined by the equation

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

Let

S

be the circle in the

x y

-plane defined by the equation

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

.

Let

P

be a point on the circle

S

with both coordinates being positive. Let the tangent to

S

P

intersect the coordinate axes at the points

M

and

N

. Then, the mid-point of the line segment

M N

must lie on the curve

Answer: (D)

Let

P = (2 cos θ, 2 sin θ)

with

θ \in (0, π /2)

.

Tangent at

P

x cos θ + y sin θ = 2

.

Intercepts:

M = (2 sec θ, 0)

N = (0, 2 csc θ)

.

Midpoint:

(h, k) = (sec θ, csc θ)

.

Since

sec^{2} θ + csc^{2} θ = \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ}

, we have

\frac{1}{h ^{2}} + \frac{1}{k ^{2}} = \frac{s i n ^{2} θ + c o s ^{2} θ}{c o s ^{2} θ s i n ^{2} θ} = \frac{1}{c o s ^{2} θ s i n ^{2} θ}

.

But

h^{2} k^{2} = sec^{2} θ csc^{2} θ = \frac{1}{c o s ^{2} θ s i n ^{2} θ}

.

So

\frac{1}{h ^{2}} + \frac{1}{k ^{2}} = h^{2} k^{2}

, giving

k^{2} + h^{2} = h^{2} k^{2}

.

Locus:

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

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Single correct option