# Find the equation of the normal to the hyperbola x^2 - 3y^2 = 144 at the positive end of the latus rectum.

## Question

Find the equation of the normal to the hyperbola $x^2 - 3y^2 = 144$ at the positive end of the latus rectum.

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## Answer

To find the equation of the normal to the hyperbola ( x^2 - 3 y^2 = 144 ) at the positive end of the latus rectum, follow these steps:

**Convert the Hyperbola to Standard Form:**[ x^2 - 3 y^2 = 144 \implies \frac{x^2}{12^2} - \frac{y^2}{4 \sqrt{3}^2} = 1 ] From this, we identify ( a = 12 ) and ( b = 4 \sqrt{3} ).**Calculate the Eccentricity (( e )):**[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{(4 \sqrt{3})^2}{12^2}} = \sqrt{1 + \frac{48}{144}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} ]**Find the Coordinates of the Positive End of the Latus Rectum:**[ \text{Latus rectum: } x = a e = 12 \times \frac{2}{\sqrt{3}} = 8 \sqrt{3} ] Substitute ( x = 8 \sqrt{3} ) back into the hyperbola equation to find ( y ): [ \frac{(8 \sqrt{3})^2}{12^2} - \frac{y^2}{4 \sqrt{3}^2} = 1 ] Simplify: [ \frac{192}{144} - \frac{y^2}{48} = 1 \implies \frac{4}{3} - \frac{y^2}{48} = 1 \implies \frac{-y^2}{48} = 1 - \frac{4}{3} = -\frac{1}{3} ] [ y^2 = 16 \implies y = \pm 4 ] So, the positive end of the latus rectum is ( (8 \sqrt{3}, 4) ).**Equation of the Normal:**The general equation for the normal to a hyperbola at ( (x_1, y_1) ) is: [ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 ] Substituting ( a = 12 ), ( b = 4 \sqrt{3} ), ( x_1 = 8 \sqrt{3} ), and ( y_1 = 4 ), we get: [ \frac{12^2 x}{8 \sqrt{3}} + \frac{(4 \sqrt{3})^2 y}{4} = 144 + 48 ] Simplify: [ \frac{144 x}{8 \sqrt{3}} + 12 y = 192 \implies 6 \sqrt{3} x + 12 y = 192 ] Dividing the entire equation by 6: [ \sqrt{3} x + 2 y = 32 ]

Hence, the equation of the normal to the hyperbola ( x^2 - 3 y^2 = 144 ) at the positive end of the latus rectum is:
**[
\boxed{\sqrt{3} x + 2 y = 32}
]**

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