Question

If the straight line xcos \[\alpha\] +y sin \[\alpha\] = p touches the curve  \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] then prove that a²cos^{2 \[\alpha\]} \[-\] b²sin² \[\alpha\] = p² ?

Answer 1

\[\text { We have, } \]

\[x\cos\alpha + y\sin\alpha = p . . . . . \left( i \right)\]

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 . . . . . \left( ii \right)\]

\[\text { As, the straight line } \left( i \right) \text { touches the curve } \left( ii \right) . \]

\[\text { So, the straight line } \left( i \right) \text { is tangent to the curve } \left( ii \right) . \]

\[\text { Also, the slope of the straight line,} m = \frac{- \cos\alpha}{\sin\alpha}\]

\[\text { And, the slope of the tangent to the curve } = \frac{dy}{dx} = \frac{b^2}{a^2} \times \frac{x}{y}\]

\[\text { So,} \frac{b^2}{a^2} \times \frac{x}{y} = \frac{- \cos\alpha}{\sin\alpha}\]

\[ \Rightarrow x b^2 \sin\alpha = - y a^2 \cos\alpha\]

\[ \Rightarrow x = \frac{- y a^2 \cos\alpha}{b^2 \sin\alpha} . . . . . \left( iii \right)\]

\[\text
{ Substituting the value of x in } \left( i \right), \text { we get }\]

\[x\cos\alpha + y\sin\alpha = p\]

\[ \Rightarrow \frac{- y a^2 \cos^2 \alpha}{b^2 \sin\alpha} + y\sin\alpha = p\]

\[ \Rightarrow \frac{- y a^2 \cos^2 \alpha + y b^2 \sin\alpha}{b^2 \sin\alpha} = p\]

\[ \Rightarrow y\left( - a^2 \cos^2 \alpha + b^2 \sin\alpha \right) = p b^2 \sin\alpha\]

\[ \Rightarrow y = \frac{p b^2 \sin\alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)}\]

\[\text { So, from } \left( iii \right), \text { we get }\]

\[x = \frac{- y a^2 \cos\alpha}{b^2 \sin\alpha}\]

\[ = \frac{- y a^2 \cos\alpha}{b^2 \sin\alpha}\]

\[ = \frac{- a^2 \cos\alpha}{b^2 \sin\alpha} \times \frac{p b^2 \sin\alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)}\]

\[ \Rightarrow x = \frac{- p a^2 \cos\alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)}\]

\[\text { Substituting the values x and y in } \left( ii \right), \text { we get }\]

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]

\[ \Rightarrow \frac{1}{a^2} \times \left( \frac{- p a^2 \cos\alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)} \right)^2 - \frac{1}{b^2} \times \left( \frac{p b^2 \sin\alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)} \right)^2 = 1\]

\[ \Rightarrow \frac{p^2 a^2 \cos^2 \alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)^2} - \frac{p^2 b^2 \sin^2 \alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)^2} = 1\]

\[ \Rightarrow \frac{p^2 a^2 \cos^2 \alpha - p^2 b^2 \sin^2 \alpha}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)^2} = 1\]

\[ \Rightarrow \frac{p^2 \left( a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \right)}{\left( b^2 \sin\alpha - a^2 \cos^2 \alpha \right)^2} = 1\]

\[ \Rightarrow \frac{p^2 \left( a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \right)}{\left( a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \right)^2} = 1\]

\[ \Rightarrow \frac{p^2}{\left( a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \right)} = 1\]

\[ \therefore p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha\]