Question

Tangent at any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ cut the axis at A and B respectively. If the rectangle OAPB (where O is origin) is completed then locus of point P is given by

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

Lets draw the diagram with given hyperbola and its tangent

Here APBO forms a rectangle. Since A and B represents point on x And y axis.

$p\equiv$ (x coordinate of A, y coordinate of B) $\equiv$ (h. k)

Then $A\equiv (h,0)$

$B\equiv (0,k)$

The given hyperbola is,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Whose tangent ig given by,

$y=mx\pm \sqrt{a^2{m}^2-b^2}$

This passes through (h,0)

$0=mh\pm \sqrt{a^2{m}^2-b^2}$

$mh=\sqrt{a^2{m}^2-b^2}$

$m^2{h}^2=a^2{m}^2-b^2$

$m^2=\frac{b^2}{a^2-h^2}$ - - - - - - -(1)

Tangent also passes through (0,k)

$k=\sqrt{a^2{m}^2-b^2}$

$\frac{k^2+b^2}{a^2}=m^2$ - - - - - -(2)

(1) and (2) $\Rightarrow$

$\frac{b^2}{a^2-h^2}=\frac{k^2+b^2}{a^2}=m^2$

$a^2{b}^2=a^2{k}^2+a^2{b}^2-h^2{k}^2-h^2{b}^2$

dividing throughout by $h^2{k}^2$

$o=\frac{a^2}{h^2}-1-\frac{b^2}{k^2}$

i.e.,$\frac{a^2}{h^2}-\frac{b^2}{k^2}$

Since (h,k) gives point P,the locus can be given as

$\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$