If the straight line `xcosalpha+ysinalpha=p` touches the curve `(x^2)/(a^2)+(y^2)/(b^2)=1` , then prove that `a^2cos^2alpha+b^2sin^2alpha=p^2dot`

Explanation

Given, line is `x cosalpha+ysin alpha=p` ……………….(i)
and curve is `x^(2)/a^(2)+y^(2)/b^(2)=1`
`rArr b^(2)x^(2)+a^(2)y^(2)=a^(2)b^(2)`……….(ii)
Now, differentiating Eq. (ii), w.r.t. x, we get
`b^(2).2x+a^(2).2y.(dy)/(dx)=-(xb^(2))/(ya^(2))` ............(iii)
From Eq.(i), `ysinalpha=p-xcosalpha`
`rArr y=-xcotalpha+p/(sinalpha)`
Thus, slope of the line is `(-cotalpha)`.
So, the given equation of line will tangent to the Eq.(ii), if `(-x/y.b^(2)/a^(2))=(-cotalpha)`
`rArr x/(a^(2)cosalpha)=y/(b^(2)sinalpha)=k` [say]
`rArr x=ka^(2)cosalpha`
and `y=b^(2)ksinalpha`
So, the line `xcosalpha+ysinalpha=p` will touch the curve `x^(2)/a^(2)+y^(2)/b^(2)` at point `(ka^(2)cosalpha, kb^(2)sinalpha)`.
From Eq.(i), `ka^(2)cos^(2)alpha+kb^(2)sin^(2)alpha=p`
`rArr a^(2)cos^(2)alpha+b^(2)sin^(2)alpha=p/k`
`rArr (a^(2)cos^(2)alpha+b^(2)sin^(2)alpha)^(2)=1`
`rArr (a^(2)cos^(2)alpha+b^(2)sin^(2)alpha)=1/k^(2)`
On dividing Eq. (iv), by Eq. (v), we get
`a^(2)cos^(2)alpha+b^(2)sin^(2)alpha=p^(2)` Hence proved.