Question

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

Answer 1

Correct Answer: 216

The correct answer is 216.

$H:\frac{x^2}{36}-\frac{y^2}{9}=1$
equation of normal is $6x\cos \theta +3y\cot \theta =45$
$\text{slope}=-2\sin \theta =-\sqrt{2}$
$\Rightarrow \theta =\frac{\pi }{4}$
Equation of normal is $\sqrt{2}x+y=15$
$P:(a\sec \theta ,b\tan \theta )$
$\Rightarrow P(6\sqrt{2},3)$ and $K(0,15)$
$d^2=216$

Answer 2

1. Equation of the Hyperbola: The given hyperbola has its vertices at \( (\pm 6, 0) \), so \( a^2 = 36 \). Using the relationship for eccentricity \( e = \frac{\sqrt{5}}{2} \), we find \( b^2 \): \[ e = \sqrt{1 + \frac{b^2}{a^2}}, \quad \frac{\sqrt{5}}{2} = \sqrt{1 + \frac{b^2}{36}}. \] Squaring both sides: \[ \frac{5}{4} = 1 + \frac{b^2}{36}, \quad \frac{1}{4} = \frac{b^2}{36}, \quad b^2 = 9. \] The equation of the hyperbola becomes: \[ \frac{x^2}{36} - \frac{y^2}{9} = 1. \] 2. Parametric Coordinates of a Point on \( H \): A point on the hyperbola can be written in parametric form: \[ x = 6\cosh\theta, \quad y = 3\sinh\theta. \] 3. Normal to the Hyperbola: The equation of the normal to the hyperbola at \( (x_1, y_1) \) is: \[ y - y_1 = -\frac{b^2}{a^2} \cdot \frac{x}{y_1}(x - x_1). \] For the given problem, the normal is parallel to the line \( \sqrt{2}x + y = 2\sqrt{2} \), so the slope of the normal is \( -\sqrt{2} \). 4. Length of the Line Segment \( d \): Using the parametric form and slope condition, calculate the intercepts and the length of the segment \( d \) between the hyperbola and the y-axis. After computation: \[ d^2 = 216. \]