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Foot of normals drawn from the point p(h,k) to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}  will always lie on the conic

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Equation of normal at any point \mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right)} is  \mathrm{P(h,k)} then \mathrm{\mathrm{y}_1\left(\mathrm{~h}-\mathrm{x}_1\right) \mathrm{a}^2+\mathrm{x}_1 \mathrm{~b}^2\left(\mathrm{k}-\mathrm{y}_1\right)=0}.Thus  \mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right)} lie on the conic \mathrm{a^2 y h-x y\left(a^2+b^2\right)+b^2 x k=0}

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HARSH KANKARIA

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