Question

Find the common tangent to the hyperbola x216y29=1 and an ellipse x24+y23=1.

Answer 1

Tangent at (psecθ,qtanϕ) for first hyerbola

xsecϕp−ytanϕq=1------(1)



Similarly at (qtanθ,psecθ) for second hyperbola
ysecθp−xtanθq=1------(2)




If (1) and (2) are common tangent then they should be identical hence

secθp=−tanϕq

secθ=−ptanϕq------(3)
−tanθq=secθp

tanθ=−secϕp


tanθ=−−qpsecϕ------(4)


sec2θ−tan2θ=1

p2q2tan2ϕ−q2p2sec2ϕ=1

p2q2tan2ϕ−q2p2(1+tan2ϕ)=1

tan2ϕ=q2p2−q2

sec2ϕ=p2p2−q2



Hence point of contact are (±p2p2−q2−−−−−√,

±q2p2−q2−−−−−√),(±q2p2−q2−−−−−√,

±p2p2−q2−−−−−√)


Length of common tangent is 2–√(p2+q2)p2−q2

Equation of common tangent is
±xp2−q2−−−−−√∓yp2−q2−−−−−√=1

x∓y=±p2−q2−−−−−−√