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Locus of perpendicular from center upon normal to the hyperbola (x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1 is

Answer» <html><body><p>`(x^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)-+y^(2))^(2)((a^(2))/(x^(2))+(b^(2))/(y^(2))) =(a^(2)-b^(2))^(2)`<br/>`(x^(2)+y^(2))^(2)((a^(2))/(x^(2))-(b^(2))/(y^(2)))=(a^(2)+b^(2))^(2)`<br/>`(x^(2)+y^(2))^(2) ((x^(2))/(a^(2))-(y^(2))/(b^(2))) =(a^(2)+b^(2))^(2)`<br/>None of these </p>Solution :Let the <a href="https://interviewquestions.tuteehub.com/tag/foot-994954" style="font-weight:bold;" target="_blank" title="Click to know more about FOOT">FOOT</a> of perpendicular from center upon any normal be `P(h,k)` Then <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> of normal to hyperbola is <br/> `(y-k) =- (h)/(k) (x-h)` <br/> or `<a href="https://interviewquestions.tuteehub.com/tag/hx-492904" style="font-weight:bold;" target="_blank" title="Click to know more about HX">HX</a> + ky = h^(2) +k^(2)` (1) <br/> Also normal at any point `<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>(a sec theta, tan theta)` on the hyperbola is <br/> `(ax)/(sec theta) +(by)/(tan theta) = a^(2) + b^(2)` (2) <br/> Comparing ratio coefficients of equations (1) and (2), <br/> We get `((a)/(sec theta))/(h)=((b)/(tan theta))/(k) =(a^(2)+b^(2))/(h^(2)+k^(2))` <br/> `rArr sec theta = (a(h^(2)+k^(2)))/(h(a^(2)+b^(2)))` and `tan theta = (b(h^(2)+k^(2)))/(k(a^(2)+b^(2)))` <br/> Squaring and subtracting `(x^(2)+y^(2))^(2) ((a^(2))/(x^(2))-(b^(2))/(y^(2))) = (a^(2)+b^(2))^(2)`</body></html>


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