Two tangents are drawn from a point on hyperbola `x^(2)-y^(2)=5` to the ellipse `(x^(2))/(9)+(y^(2))/(4)=1`. If they make angle `alpha and beta` with

Question

Two tangents are drawn from a point on hyperbola `x^(2)-y^(2)=5` to the ellipse `(x^(2))/(9)+(y^(2))/(4)=1`. If they make angle `alpha and beta` with

Two tangents are drawn from a point on hyperbola `x^(2)-y^(2)=5` to the ellipse `(x^(2))/(9)+(y^(2))/(4)=1`. If they make angle `alpha and beta` with x-axis, then
A. `alpha-beta=pm(pi)/(2)`
B. `alpha+beta=(pi)/(2)`
C. `alpha+beta=pi`
D. `alpha+beta=0`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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

Correct Answer - B
Equation of tangent to ellipse `(x^(2))/(9)+(y^(2))/(4)=1` having slope m is
`y=mx+sqrt(9m^(2)+4)`.
Since it passes through point `(sqrt5 sec theta, sqrt5 tan theta)` on the hyperbola, we gave
`sqrt5 tantheta=sqrt5 sectheta+sqrt(9m^(2)+4)`
`therefore" "(sqrt5 tan theta-m sqrt5 sec theta)^(2)=9m^(2)+4`
`rArr" "(9-5 sec^(2)theta)m^(2)+10sec theta tan thetam +4-5 tan^(2)theta=0`
This equation has roots `m_(1) and m_(2)`, which are slopes of tangents.
Now, `m_(1)m_(2)=(4-5 tan^(2)theta)/(9-5sec^(2)theta)=(4-5(sec^(2)theta-1))/(9-5sec^(2)theta)=1`
`therefore" (tan alpha)(tan beta)=1`
`rArr" "alpha+beta=(pi)/(2)`