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Hyperbola

Section: Mathematics

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Question : 2 of 34

Marks: +1, -0

If the normal drawn to the hyperbola

xy=16

(8,2)

meets the hyperbola again at a point

(\alpha, \beta)

|\beta|+\;\frac{1}{|\alpha|}=

[AP EAPCET 24 May 2025 S1]

$40$
$34$
$28$
$54$

Solution:

Given hyperbola equation is

x y=16

Differentiate it w.r.t

x

, we get

\;y+x \cdot \;\frac{dy}{dx}\;=0

\Rightarrow\;\;\frac{dy}{dx}\;=\;\frac{-y}{x}

Now, slope of tangent at

(8,2)

is

m_1=\;\frac{-2}{8}=\;\frac{-1}{4}

And slope of normal is

m_n=\;\frac{-1}{-\frac{1}{4}}=4

Now, equation of normal at

(8,2)

is

\;y-2=4(x-8)

\Rightarrow\;\; y=4x-30

Normal intersects the hyperbola again at a point

(\alpha, \beta)

,

So,

\alpha \beta=16

and

\beta=4\alpha-30

Solving these two equations, we get

\;\alpha(4\alpha-30)=16

\;\Rightarrow\;\; 2\alpha^2-15\alpha-8=0

\text{So, }\alpha\;=\;\frac{-(-15) \pm \sqrt{(-15)^2-4 \cdot 2 \cdot (-8)}}{2 \cdot 2}

\;=\;\frac{15 \pm \sqrt{289}}{4}=\;\frac{15 \pm 17}{4}

So,

\alpha_1=\;\frac{15+17}{4}=\;\frac{32}{4}=8

and

\alpha_2=\;\frac{15-17}{4}=\;\frac{-1}{2}

Since

(8,2)

is one point, the other point is where

\alpha=\;\frac{-1}{2}

So,

\beta=4\alpha-30

\;=4\left(\frac{-1}{2}\right)-30=-32

\;\therefore |\beta|+\;\frac{1}{|\alpha|}=|-32|+\;\frac{1}{|-1/2|}

\;\; =32+2=34

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