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  • Locus of foot of the perpendicular from centre upon normal to the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

Locus of foot of the perpendicular from centre upon normal to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} is

Option: 1

\mathrm{\left(x^2+y^2\right)^2\left(\frac{a^2}{x^2}+\frac{b^2}{y^2}\right)=\left(a^2-b^2\right)^2}


Option: 2

\mathrm{\left(x^2+y^2\right)^2\left(\frac{a^2}{x^2}-\frac{b^2}{y^2}\right)=\left(a^2+b^2\right)^2}


Option: 3

\mathrm{\left(x^2+y^2\right)^2\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)=\left(a^2+b^2\right)^2}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

Let the foot of perpendicular from centre upon any normal be P(h, k).

Then, equation of normal passing through P(h, k) to hyperbola is

\mathrm{ (y-k)=-\frac{h}{k}(x-h) \Rightarrow h x+k y=h^2+k^2 }             ....(i)
Also, normal at any point \mathrm{R(a \sec \theta, b \tan \theta)} on the hyperbola is

\mathrm{ \frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2+b^2 }                                                       ....(ii)

Comparing coefficients of equations (i) and (ii), we get

\mathrm{ \begin{aligned} & \frac{a}{\sec \theta}=\frac{b}{\tan \theta}=\frac{a^2+b^2}{h^2+k^2} \\\\ & \Rightarrow \sec \theta=\frac{a\left(h^2+k^2\right)}{h\left(a^2+b^2\right)} \text { and } \tan \theta=\frac{b\left(h^2+k^2\right)}{k\left(a^2+b^2\right)} \end{aligned} }

Squaring and subtracting, we get

\mathrm{ \left(h^2+k^2\right)^2\left(\frac{a^2}{h^2}-\frac{b^2}{k^2}\right)=\left(a^2+b^2\right)^2 }
Hence, the locus of (h, k) is

\mathrm{ \left(x^2+y^2\right)^2\left(\frac{a^2}{x^2}-\frac{b^2}{y^2}\right)=\left(a^2+b^2\right)^2 }

Posted by

Gautam harsolia

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