# Examples on Tangents and Chords of Hyperbolas Set 1

Go back to  'Hyperbola'

Example - 22

Find the locus of the mid-points of the chords of the hyperbola  \begin{align}\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\end{align} which subtend a right angle at the origin O.

Solution: Assume  $$P(h,\,k)$$ to be the mid-point of any such chord which subtends a right angle at the origin O.

The equation of this chord is

\begin{align}&\qquad\;\;\ T(h,\,k)=S(h,\,k) \\ & \Rightarrow \quad \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}}=\frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 1 \right) \\\end{align}

Now, suppose this chord intersects the hyperbola in the points OQ and OR. We can write the joint equation of OQ and OR by homogenizing the equation of the hyperbola using the equation of the chord obtained in (1):

\begin{align}&\qquad\;\; \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}={{\left\{ \frac{\left(\begin{align} \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}}\end{align} \right)}{\left( \begin{align}\frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}}\end{align} \right)} \right\}}^{2}} \\& \Rightarrow\quad \left( \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}} \right){{\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}} \right)}^{2}}={{\left( \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}} \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 2 \right) \end{align}

OQ and OR will be perpendicular if in (2) ,

\begin{align}&\qquad\;\;\text{Coeff}\text{. of }{{x}^{2}}+\text{Coeff}\text{. of }{{y}^{2}}=0 \\\\ & \Rightarrow\quad\frac{1}{{{a}^{2}}}{{\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}} \right)}^{2}}-\frac{1}{{{b}^{2}}}{{\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}} \right)}^{2}}-\frac{{{h}^{2}}}{{{a}^{4}}}-\frac{{{k}^{2}}}{{{b}^{4}}}=0 \\ \end{align}

Thus, the locus of mid point P is

$\left( \frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}} \right){{\left( \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}} \right)}^{2}}=\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{4}}}$

Example - 23

Find the locus of the point of intersection of tangents drawn at the extremities of a normal chord of the hyperbola\begin{align}\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\end{align}

Solution: Let QR be a normal chord of the given hyperbola and let the tangents at Q and R meet at$$P(h,\,k).$$

QR is basically the chord of contact of the tangents drawn from  $$P(h,\,k)$$ and thus in terms of h and k, the equation of QR will be

\begin{align}&\qquad\;\;\;\ T(h,\,k)=0 \\ & \Rightarrow\quad \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 1 \right) \\\end{align}

Also, since QR is a normal chord, there must be some $$\theta$$ for which the equation of QR can be written as

$ax\cos \theta +by\cot \theta ={{a}^{2}}+{{b}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 2 \right)$

From (1) and (2), we have

\begin{align}&\qquad\;\;\frac{h/{{a}^{2}}}{a\cos \theta }=\frac{-k/{{b}^{2}}}{b\cot \theta }=\frac{1}{{{a}^{2}}+{{b}^{2}}} \\&\Rightarrow\quad \sec \theta =\frac{{{a}^{3}}}{h({{a}^{2}}+{{b}^{2}})},\,\,\,\,\,\tan \theta =\frac{-{{b}^{3}}}{k({{a}^{2}}+{{b}^{2}})} \\\end{align}

From  $${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1,$$ we have

$\frac{{{a}^{6}}}{{{h}^{2}}}-\frac{{{b}^{6}}}{{{k}^{2}}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$

Thus, the locus of P is

$\frac{{{a}^{6}}}{{{x}^{2}}}-\frac{{{b}^{6}}}{{{y}^{2}}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$

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