The locus of the foot of the perpendicular from the center of the hyperbola `x y=1` on a variable tangent is `(x^2-y^2)=4x y` (b) `(x^2-y^2)=1/9` `(x^

Question

The locus of the foot of the perpendicular from the center of the hyperbola `x y=1` on a variable tangent is `(x^2-y^2)=4x y` (b) `(x^2-y^2)=1/9` `(x^

The locus of the foot of the perpendicular from the center of the hyperbola `x y=1` on a variable tangent is `(x^2-y^2)=4x y` (b) `(x^2-y^2)=1/9` `(x^2-y^2)=7/(144)` (d) `(x^2-y^2)=1/(16)`
A. `(x^(2)-y^(2))^(2)=4xy`
B. `(x^(2)+y^(2))^(2)=2xy`
C. `(x^(2)+y^(2))=4xy`
D. `(x^(2)+y^(2))^(2)=4xy`

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
Let the foot of perpenicular from O(0, 0) to the hyperbola be `P(h,k)`
Slope of `OP=(k)/(h)`
Then the equation of tangent to the hypebola is
`y-k=-(h)/(k)(x-k)` ltBrgt `"or "hx+ky=h^(2)+k^(2)`
Solving it with xy = 1, we have
`hx+(k)/(x)=h^(2)+k^(2)`
`"or "hx^(2)-(h^(2)+k^(2))x+k=0`
This equation must have real and equal roots. Hence,
D = 0
`"or "(h^(2)+k^(2))^(2)-4hk=0`
`"or "(x^(2)+y^(2))^(2)=4xy`