A normal to the hyperbola 4x^2 - 9y^2 = 36 me | vedclass

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(C) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$.Let the point on the hyperbola be

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$9x^2 - 4y^2 = 169$

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(C) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$.Let the point on the hyperbola be $(x_0, y_0)$. The equation of the normal at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$.Here $a^2 = 9$ and $b^2 = 4$,so the equation is $\frac{9x}{x_0} + \frac{4y}{y_0} = 9 + 4 = 13$.This normal meets the $x$-axis at $A(\frac{13x_0}{9}, 0)$ and the $y$-axis at $B(0, \frac{13y_0}{4})$.Since $OABP$ is a parallelogram with $O(0,0)$,the coordinates of $P(x, y)$ are given by $P = A + B = (\frac{13x_0}{9}, \frac{13y_0}{4})$.Thus,$x = \frac{13x_0}{9} \Rightarrow x_0 = \frac{9x}{13}$ and $y = \frac{13y_0}{4} \Rightarrow y_0 = \frac{4y}{13}$.Since $(x_0, y_0)$ lies on the hyperbola $4x_0^2 - 9y_0^2 = 36$,we substitute the values:$4(\frac{9x}{13})^2 - 9(\frac{4y}{13})^2 = 36$$4 \cdot \frac{81x^2}{169} - 9 \cdot \frac{16y^2}{169} = 36$$\frac{324x^2 - 144y^2}{169} = 36$$324x^2 - 144y^2 = 36 \cdot 169$Dividing by $36$,we get $9x^2 - 4y^2 = 169$.

$A$ normal to the hyperbola $4x^2 - 9y^2 = 36$ meets the coordinate axes $x$ and $y$ at $A$ and $B$,respectively. If the parallelogram $OABP$ ($O$ being the origin) is formed,then the locus of $P$ is

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