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

1453986_765352_ans_5cdf595b262e4d5bba1852c003af79f0.png

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