First slide

Hyperbola in conic sections

Question

Normal at point (5, 3) to the rectangular hyperbola $x y - y - 2 x - 2 = 0$
meets the curve at the point whose coordinates are

Moderate

A

(0, –2)
B

(–1, 0)
C

$(\frac{1}{4}, \frac{- 10}{3})$
D

$(\frac{3}{4}, - 14) .$

Solution

Equation of the hyperbola can be written as $(x - 1) (y - 2) = 4$
or $y = 2 + \frac{4}{x - 1}$
$\frac{d y}{d x} = - \frac{4}{(x - 1)^{2}}$
Any point on the hyperbola is
$(1 + 2 t, 2 + \frac{2}{t}) = (5, 3) \Rightarrow t = 2$
So slope of the normal at (5, 3) is
$- {\frac{d x}{d y}|}_{(5, 3)} = 4$
Equation of the normal at (5, 3) is y – 3 = 4(x – 5) which will pass through
$(1 + 2 t, 2 + \frac{2}{t}) if 2 + \frac{2}{t} = 4 (1 + 2 t) - 17$
$\Rightarrow 8 t^{2} - 15 t - 2 = 0 \Rightarrow t = 2, - \frac{1}{8}$ .
For $t = - \frac{1}{8},$ we get the required point as
$(1 - \frac{2}{8}, 2 - \frac{2}{1 / 8}) = (\frac{3}{4}, - 14)$

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