Question

If the straight line xcos$\alpha$ + ysin$\alpha$ = p touches the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then prove that a²cos²$\alpha$ $-$ b²sin²$\alpha$ = p².

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\cos \alpha +y\sin \alpha =p.....\left(\mathrm{i}\right) \\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.....\left(\mathrm{ii}\right) \\ \mathrm{As},\mathrm{the}\mathrm{straight}\mathrm{line}\left(\mathrm{i}\right)\mathrm{touches}\mathrm{the}\mathrm{curve}\left(\mathrm{ii}\right). \\ \mathrm{So},\mathrm{the}\mathrm{straight}\mathrm{line}\left(\mathrm{i}\right)\mathrm{is}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}\left(\mathrm{ii}\right). \\ \mathrm{Also},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{straight}\mathrm{line},m=\frac{-\cos \alpha }{\sin \alpha } \\ \mathrm{And},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}=\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{b^2}{a^2}\times \frac{x}{y} \\ \mathrm{So},\frac{b^2}{a^2}\times \frac{x}{y}=\frac{-\cos \alpha }{\sin \alpha } \\ \Rightarrow x{b}^2\sin \alpha =-y{a}^2\cos \alpha \\ \Rightarrow x=\frac{-y{a}^2\cos \alpha }{b^2\sin \alpha }.....\left(\mathrm{iii}\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ x\cos \alpha +y\sin \alpha =p \\ \Rightarrow \frac{-y{a}^2{\cos }^2\alpha }{b^2\sin \alpha }+y\sin \alpha =p \\ \Rightarrow \frac{-y{a}^2{\cos }^2\alpha +y{b}^2\sin \alpha }{b^2\sin \alpha }=p \\ \Rightarrow y\left(-a^2{\cos }^2\alpha +b^2\sin \alpha \right)=p{b}^2\sin \alpha \\ \Rightarrow y=\frac{p{b}^2\sin \alpha }{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)} \\ \mathrm{So},\mathrm{from}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get} \\ x=\frac{-y{a}^2\cos \alpha }{b^2\sin \alpha } \\ =\frac{-y{a}^2\cos \alpha }{b^2\sin \alpha } \\ =\frac{-a^2\cos \alpha }{b^2\sin \alpha }\times \frac{p{b}^2\sin \alpha }{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)} \\ \Rightarrow x=\frac{-p{a}^2\cos \alpha }{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)} \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}x\mathrm{and}y\mathrm{in}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} \\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{1}{a^2}\times {\left(\frac{-p{a}^2\cos \alpha }{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}\right)}^2-\frac{1}{b^2}\times {\left(\frac{p{b}^2\sin \alpha }{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}\right)}^2=1 \\ \Rightarrow \frac{p^2{a}^2{\cos }^2\alpha }{{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}^2}-\frac{p^2{b}^2{\sin }^2\alpha }{{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}^2}=1 \\ \Rightarrow \frac{p^2{a}^2{\cos }^2\alpha -p^2{b}^2{\sin }^2\alpha }{{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}^2}=1 \\ \Rightarrow \frac{p^2\left(a^2{\cos }^2\alpha -b^2{\sin }^2\alpha \right)}{{\left(b^2\sin \alpha -a^2{\cos }^2\alpha \right)}^2}=1 \\ \Rightarrow \frac{p^2\left(a^2{\cos }^2\alpha -b^2{\sin }^2\alpha \right)}{{\left(a^2{\cos }^2\alpha -b^2{\sin }^2\alpha \right)}^2}=1 \\ \Rightarrow \frac{p^2}{\left(a^2{\cos }^2\alpha -b^2{\sin }^2\alpha \right)}=1 \\ \therefore p^2=a^2{\cos }^2\alpha -b^2{\sin }^2\alpha \end{gathered}$$