The exhaustive set of values of `alpha^2` such that there exists a tangent to the ellipse `x^2+alpha^2y^2=alpha^2` and the portion of the tangent inte

Question

The exhaustive set of values of `alpha^2` such that there exists a tangent to the ellipse `x^2+alpha^2y^2=alpha^2` and the portion of the tangent inte

The exhaustive set of values of `alpha^2` such that there exists a tangent to the ellipse `x^2+alpha^2y^2=alpha^2` and the portion of the tangent intercepted by the hyperbola `alpha^2x^2-y^2=1` subtends a right angle at the center of the curves is:

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

Equation of tangent at point `P(alpha cos theta, sin theta)` is
`(x)/(alpha)cos theta+(x)/(1)sin theta=1" (1)"`
Let it cut the hyperbola at point P and Q.
Homogenising hyperbola `alpha^(2)x^(2)-y^(2)=1` w.r.t. line (1), we get
`alpha^(2)x^(2)-y^(2)=((x)/(alpha)cos theta+y sin theta)^(2)`
This is an equation of pair of straight lines OP and OQ.
Given that `anglePOQ=90^(@)`.
`therefore" Coefficient of "x^(2)+"Coefficient of "y^(2)=0`
`rArr" "alpha^(2)-(cos^(2)theta)/(alpha^(2))-1-sin^(2)=0`
`rArr" "alpha^(2)-(cos^(2)theta)/(alpha^(2))-1-1+cos^(2)theta=0`
`rArr" "cos^(2)theta=(alpha^(2)2-alpha^(2))/(alpha-1)`
`"Now, "0lecos^(2)theta le 1`
`rArr" "0le (alpha^(2)(2-alpha^(2)))/(alpha^(2)-1)le1`
Solving, we get
`alpha^(2) in [(sqrt5+1)/(2),2]`