If any line perpendicular to the transverse axis cuts the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` and the conjugate hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=-

Question

If any line perpendicular to the transverse axis cuts the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` and the conjugate hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=-

If any line perpendicular to the transverse axis cuts the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` and the conjugate hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=-1` at points `Pa n dQ` , respectively, then prove that normal at `Pa n dQ` meet on the x-axis.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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

Let the perpendicular line cuts the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
at point `P(x_(1),y_(1))` and the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`
at point `Q(x_(1),y_(1))`.
Normal to the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
at point P is
`(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)+b^(2)" (1)"`
Normal to the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`
at Q is
`(a^(2)x)/(x_(1))+(b^(2)y)/(y_(2))=a^(2)+b^(2)" (2)"`
In (1) and (2), putting y = 0, we get
`x=(a^(2)+b^(2))/(a^(2))x_(1)`
Hence, both normals meet on the x-axis.