Test Index

Hyperbola

Section: Mathematics

Question : 5 of 34

Marks: +1, -0

\theta

(1,1)

4x^2-5y^2-20=0

\tan \theta=

Solution:

Given equation of hyperbola can be written as

\frac{x^2}{5}-\frac{y^2}{4}=1

which is of the form

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \cdot\cdot\cdot\cdot\cdot\cdot\cdot \text{(i)}

where,

a=\sqrt{5}, b=2

As we know that, equation of any tangent to the hyperbola Eq. (i) having slope

m

y=m x \pm \sqrt{a^2 m^2-b^2}=m x+\sqrt{5 m^2-4} \cdot\cdot\cdot\cdot\cdot\cdot\cdot \text{(ii)}

According to question, Eq. (ii) passes through

(1,1)

So,

1=m \times 1 \pm \sqrt{5 m^2-4}

\Rightarrow 1-m=\pm \sqrt{5 m^2-4}

\Rightarrow (1-m)^2=5 m^2-4

\Rightarrow 4 m^2+2 m-5=0 \cdot\cdot\cdot\cdot\cdot\cdot\cdot \text{(iii)}

Let

m_1, m_2

slopes of these two tangents be the roots of Eq. (iii).

So,

m_1+m_2=\frac{-2}{4}=\frac{-1}{2}

m_1 m_2=\frac{-5}{4}

\therefore \tan \theta=\left| \frac{m_1-m_1}{1+m_1 m_2} \right|

=\left| \frac{\sqrt{(m_1+m_2)^2-4 m_1 m_2}}{1+m_1 m_2} \right|

=\left| \frac{\sqrt{\left(\frac{-1}{2}\right)^2-4 \times \left(\frac{-5}{4}\right)}}{1+\left(\frac{-5}{4}\right)} \right|

=\left| \frac{\sqrt{\frac{1}{4}+\frac{20}{4}}}{1-\frac{5}{4}} \right|=\left| \frac{\sqrt{21}}{2} \times \frac{4}{-1} \right|

=|-2 \sqrt{21}|=2 \sqrt{21}

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