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10-2. Parabola, Ellipse, Hyperbola
hard
यदि सरल रेखा $x\cos \alpha + y\sin \alpha = p$ अतिपरवलय $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ की स्पर्श रेखा हो, तब
A
${a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}$
B
${a^2}{\cos ^2}\alpha - {b^2}{\sin ^2}\alpha = {p^2}$
C
${a^2}{\sin ^2}\alpha + {b^2}{\cos ^2}\alpha = {p^2}$
D
${a^2}{\sin ^2}\alpha - {b^2}{\cos ^2}\alpha = {p^2}$
Solution
(b) $x\cos \alpha + y\sin \alpha = p$
$\Rightarrow y = – \cot \alpha .\,\,x + p{\rm{cosec }}\alpha $
यह अतिपरवलय $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ की स्पर्श है।
अत :${p^2}{\rm{cose}}{{\rm{c}}^2}\alpha = {a^2}{\cot ^2}\alpha – {b^2}$
$ \Rightarrow {a^2}{\cos ^2}\alpha – {b^2}{\sin ^2}\alpha = {p^2}$.
Standard 11
Mathematics
