Question

nd are inclined at avgicsTangents are drawn from the point `(alpha, beta)` to the hyperbola `3x^2- 2y^2=6` and are inclined atv angle `theta and phi` to the x-axis.If `tan theta.tan phi=2`, prove that `beta^2 = 2alpha^2 - 7`.

Answer 1

Correct Answer - 7
The given hyperbola is
`3x^(2)-2y^(2)=6`
`"or "(x^(2))/(2)-(y^(2))/(3)=1`
Equation of tangent is
`y=mx pm sqrt(a^(2)m^(2)-b^(2))`
`"or "(y-mx)^(2)=a^(2)m^(2)-b^(2)`
Tangents from the point `(alpha, beta)` will be
`(beta-malpha)^(2)=2m^(2)-3" "("Since"a^(2)=2 and b^(2)=3)`
`"or "m^(2)alpha^(2)+beta^(2)-2malphabeta-2m^(2)+3=0`
`m^(2)(alpha^(2)-2)-2alpha betam+beta^(2)+3=0`
`m_(1)*m_(2)=(beta^(2)+3)/(alpha^(2)-3)=2=tan thetatan phi`
`therefore" "beta^(2)+3=2(alpha^(2)-2)`
`"or "2alpha^(2)-beta^(2)=7`