Question

If the straight line x cos $\alpha$ + y sin $\alpha$ = p touches the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then prove that $a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^2$.

Answer 1

Given, line is $xcos\alpha +ysin\alpha =p$
and curve is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow b^2{x}^2+a^2{y}^2=a^2{b}^2$
Now, differentiating Eq. (ii) w.r.t. x we get
$b^2.2x+a^2.2y.\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-2b^{2}x}{2a^{2}y}=\frac{-x{b}^2}{y{a}^2}$
From Eq. (i) . $ysin\alpha =p-xcos\alpha \Rightarrow y=-xcot\alpha +\frac{p}{sin\alpha }$
Thus, slope of the line is $(-cot\alpha )$.
So, the given equation of line will be tangent to the Eq. (ii), if $(-\frac{x}{y}.\frac{b^2}{a^2})=(-cot\alpha )\Rightarrow \frac{x}{a^{2}cos\alpha }=\frac{y}{b^{2}sin\alpha }=k\Rightarrow x=k{a}^{2}cos\alpha andy=b^{2}ksin\alpha$
So the line x cos $\alpha$ + y sin $\alpha$ = p will touch the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ at the point $(k{a}^{2}cos\alpha ,k{b}^{2}sin\alpha )$
From Eq. (i) . $k{a}^{2}co{s}^2\alpha +k{b}^{2}si{n}^2\alpha =p\Rightarrow a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =\frac{\rho }{k}\Rightarrow (a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha {)}^2=\frac{{\rho }^2}{k^2}$
From Eq. (ii), $b^2{k}^2{a}^{4}co{s}^2\alpha +a^2{k}^2{b}^{4}si{n}^2\alpha =a^2{b}^2\Rightarrow k^2(a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha )=1\Rightarrow (a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha )=\frac{1}{k^2}$
On dividing Eq. (iv) by Eq. (v) . we get
$a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^2$