दीर्घवृत्त frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}} = 1 के केन?

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10-2. Parabola, Ellipse, Hyperbola

hard

दीर्घवृत्त $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ के केन्द्र से इसकी किसी स्पर्श रेखा पर डाले गये लम्ब के पाद का बिन्दुपथ है

${({x^2} + {y^2})^2} = {b^2}{x^2} + {a^2}{y^2}$

${({x^2} + {y^2})^2} = {b^2}{x^2} - {a^2}{y^2}$

${({x^2} + {y^2})^2} = {a^2}{x^2} - {b^2}{y^2}$

${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$

Solution

(d) $\frac{{x\cos \theta }}{a} + \frac{{y\sin \theta }}{b} = 1,$ ढाल या प्रवणता $ \equiv – \frac{{b\cot \theta }}{a} \times \frac{k}{h} = – 1$
$\frac{{h\cos \theta }}{a} + \frac{{k\sin \theta }}{b} = 1$
चूँकि ${\rm{cosec}}\theta = \sqrt {1 + \frac{{{a^2}{h^2}}}{{{k^2}{b^2}}}} $==> $\frac{{{h^2}}}{{kb}} + \frac{k}{b} = {\rm{cosec}}\theta $
==> $({h^2} + {k^2}) = bk\,{\rm{cosec}}\theta $ = $\frac{{bk(\sqrt {{k^2}{b^2} + {a^2}{h^2}} )}}{{bk}}$
==> ${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$.

Standard 11

Mathematics