The vertices of a hyperbola H are ( 6 , 0 ) and its eccentricity is 5 2 . Let N be the normal to H at a point in the first quadrant and parallel to the line 2 ? x + y = 2 2 . If d is the length of the line segment of N between H and the y-axis, then

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Solution

Q.
The vertices of a hyperbola H are $(\pm 6, 0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$ . Let N be the normal to H at a point in the first quadrant and parallel to the line $\sqrt{2} x + y = 2 \sqrt{2}$ . If $d$ is the length of the line segment of N between H and the y-axis, then $d^{2}$ is equal to _________ . [2023]

Ans.
(216)

$Equation of hyperbola = \frac{x^{2}}{36} - \frac{y^{2}}{b^{2}} = 1$

$Let H be (x_{1}, y_{1})$

$b^{2} = 36 (\frac{5}{4} - 1) = 9$

$Hyperbola: \frac{x^{2}}{36} - \frac{y^{2}}{9} = 1$

$Equation of normal: \frac{36 x}{x_{1}} + \frac{9 y}{y_{1}} = 45$ $\Rightarrow - \frac{4 y_{1}}{x_{1}} = - \sqrt{2}$

$x_{1} = 2 \sqrt{2} y_{1}$ and $\frac{{(2 \sqrt{2} y_{1})}^{2}}{36} - \frac{y_{1}^{2}}{9} = 1$

$\frac{8 y_{1}^{2}}{36} - \frac{y_{1}^{2}}{9} = 1$ $\Rightarrow y_{1} = 3 (y_{1} > 0)$ $\Rightarrow x_{1} = 6 \sqrt{2}$

$Equation of normal: y = - \sqrt{2} x + 15$

$This intersects the y–axis at (0, 15)$

$d^{2} = {(6 \sqrt{2} - 0)}^{2} + {(3 - 15)}^{2}$

$= 72 + 144 = 216$

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