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Let S be the circle in the x y -plane de...

Let `S` be the circle in the `x y` -plane defined by the equation `x^2+y^2=4.` (For Ques. No 15 and 16) 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 (a)`(x+y)^2=3x y` (b) `x^(2//3)+y^(2//3)=2^(4//3)` (c) `x^2+y^2=2x y` (d) `x^2+y^2=x^2y^2`

A

`(x-4)^(2)=3xy`

B

`x^(23)+y^(23)=2^(4//3)`

C

`x^(2)+y^(2)=2xy`

D

`x^(2)+y^(2)=x^(2)y^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
We have
`x^(2)+y^(2)=4`
Let `P(2 cos theta, 2sintheta)` be a point on a circle.
`therefore` Tangent at P is
` 2costhetax+2sinthetay=4`
`rArr x costheta+ysintheta=2`

`therefore` The coordinates at `M((2)/(costheta),0)) and N((0,(2)/(sintheta))`
Let (h,k) is mid point of MN
`thereforeh=(1)/(costheta)and k=(1)/(sintheta)`
`rArrcostheta=(1)/(h)and sintheta=(1)/(k)`
`rArrcos^(2)theta+sin^(2)theta=(1)/(h^(2))+(1)/(k^(2))rArr1= (h^(2)+k^(2))/(h^(2)+k^(2))`
`rArrh^(2)+k^(2)=h^(2)k^(2)`
`therefore" Mid-point of MN lie on the curve " x^(2)+y^(2)=x^(2)y^(2)`
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