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x^(2)/a^2+y^2/b^2=1...

`x^(2)/a^2+y^2/b^2=1`

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The correct Answer is:
`(-b^2x)/(a^2y)`
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Statement 1 : If from any point P(x_1, y_1) on the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=-1 , tangents are drawn to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1, then the corresponding chord of contact lies on an other branch of the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=-1 Statement 2 : From any point outside the hyperbola, two tangents can be drawn to the hyperbola.

If a point (x_(1),y_(1)) lies in the shaded region (x^(2))/(a^(2))-(y^(2))/(b^(2))=1, shown in the figure,then (x^(2))/(a^(2))-(y^(2))/(b^(2))<0 statement 2: If P(x_(1),y_(1)) lies outside the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1, then (x_(1)^(2))/(a^(2))-(y_(1)^(2))/(b^(2))<1

The asymptotes of the hyperbola (x^(2))/(a_(1)^(2))-(y^(2))/(b_(1)^(2))=1 and (x^(2))/(a_(2)^(2))-(y^(2))/(b_(2)^(2))=1 are perpendicular to each other. Then,