Question

Equation of tangent for a point at hyperbola

Answer 1

down voteaccepted(1a)Let y=mx+cy=mx+c be a tangent of x2a2−y2b2=1x2a2−y2b2=1 Let (h,k)(h,k) be the point of intersection.So, k=mh+c,b2h2−a2k2=a2b2⟹b2h2−a2(mh+c)2=a2b2k=mh+c,b2h2−a2k2=a2b2⟹b2h2−a2(mh+c)2=a2b2 or,h2(b2−a2m2)−2a2mch−a2b2−a2c2=0h2(b2−a2m2)−2a2mch−a2b2−a2c2=0 This is a quadratic equation in h,h, for tangency, the roots need to be same, to make the two points of intersection coincident.⟹(−2a2mc)2=4⋅(b2−a2m2)(−a2b2−a2c2)⟹(−2a2mc)2=4⋅(b2−a2m2)(−a2b2−a2c2) a2m2c2=a2m2b2+a2m2c2−b4−b2c2a2m2c2=a2m2b2+a2m2c2−b4−b2c2 cancelling out a2a2 as a≠0a≠0 0=a2m2b2−b4−b2c20=a2m2b2−b4−b2c2 0=a2m2−b2−c20=a2m2−b2−c2 cancelling out b2b2 as b≠0b≠0 ⟹c2=a2m2−b2⟹c=±a2m2−b2−−−−−−−−√⟹c2=a2m2−b2⟹c=±a2m2−b2 (1b) Alternatively, we can take the central conic to be Ax2+By2=1Ax2+By2=1 Applying the same method we get, ABc2=A+m2BABc2=A+m2B Here A=1a2,B=−1b2⟹c2=a2m2−b2A=1a2,B=−1b2⟹c2=a2m2−b2 (2a)Using calculus as André Nicolas has already done, we find the equation of the tangent to be xx1a2−yy1b2−1=0xx1a2−yy1b2−1=0 Comparing with mx−y+c=0mx−y+c=0 we get, x1ma2=y1b2=−1cx1ma2=y1b2=−1c Now (x1,y1)(x1,y1) lies on the given curve, so x21a2−y21b2=1x12a2−y12b2=1 , eliminating x1,y1x1,y1 , we shall get the desired result.(2b) we know , the parametric equation of the given curve is x=asect,y=btantx=asec⁡t,y=btan⁡t So, the equation of the tangent becomes xsecta−ytantb−1=0xsec⁡ta−ytan⁡tb−1=0 Comparing with mx−y+c=0mx−y+c=0 we get, sectma=tantb=−1csec⁡tma=tan⁡tb=−1c So, sect=−mac,tant=−bcsec⁡t=−mac,tan⁡t=−bc Now use eliminate tt .