Question

Question #53c2f

Answer 1

`sqrt2x-y-1=0.`

Explanation:

Eqn. of Hyperbola : `x^2-y^2=1.`

Diff. both sides w.r.t. `x, 2x-2ydy/dx=0.`
`:. dy/dx=x/y, y!=0.`
`:.[dy/dx]_(x=sqrt2,y=1) = sqrt2.`
`:.` Slope of tgt. to Hyp. at pt. `(sqrt2,1)=sqrt2,` & pt. `(sqrt2,1)` is on the tgt.

Hence, reqd. eqn. of tgt. is `: y-1=sqrt2*(x-sqrt2),` writing it in std. form, `sqrt2x-y-1=0.`

Here, we had to find the eqn. of tgt. using implicit diffn. Otherwise, eqn. of the tgt. line at a pt.`(h,k)` on Hyp. : `x^2-y^2=1,` is, `hx-ky=1.`

Here, pt. `(h,k)=(sqrt2,1)` is on Hyp. `: x^2-y^2=1,` since it satisfy the eqn. of Hyp.So, tgt.line is `sqrt2x-y=1,` as before!