Question

# The normal to the hyperbola $$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$ meets the axes in $$M$$ and $$N$$ and the lines $$MP$$ and $$NP$$ are drawn at right angles to the axes. Prove that the locus of $$P$$ is the hyperbola $${ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ y }^{ 2 }={ \left( { a }^{ 2 }+{ b }^{ 2 } \right) }^{ 2 }$$

Solution

## Normal to hyperbola at $$(x_1 , y_1)$$ on hyperbola is $$a^2 xx_1 +b^2y/y_1 =a^2 e^2$$$$M(x_1e^2, 0)$$$$N\left (O, \dfrac {a^2e^2}{b^2} y_1\right)$$$$P\left(x_1 e^2 , \dfrac {a^2e^2}{b^2}y_1 \right)$$$$x=x_1 e_1^2, y=\dfrac {a^2e^2}{b^2}y_1$$$$\Rightarrow \ x_1 =\dfrac {x}{e_1^2},\ y_1=\dfrac {a^2 b^2}{a^2 b^2}$$$$\Rightarrow \ x_1=\dfrac {a\ a^2}{a^2 +b^2}, y_1 =\dfrac {a^2 b^2}{a^2 +b^2}$$$$\dfrac {x_1^2}{a^2}-\dfrac {y_1^2}{b^2}=1 \Rightarrow \dfrac {a^4 x^2}{(a^2 +b^2)^2 a^2}-\dfrac {y^2 b^4}{(a^2 +b^2)^2 b^2}=1$$$$\Rightarrow \ a^2x^2-b^2 y^2 =(a^2 +b^2)^2$$Maths

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