Basic Examples on Hyperbolas Set 3

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Example - 7

Find the equation of the chord joining the points \(P({\theta _1})\) and \(Q({\theta _2})\) on the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}\)

Solution : The two points P and Q have the coordinates \((a\sec {\theta _1},\,\,b\tan {\theta _1})\)and \((a\sec {\theta _2},\,\,b\tan {\theta _2})\) respectively. Using the two-point form, the equation of PQ is

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\frac{{y - b\tan {\theta _1}}}{{x - a\sec {\theta _1}}} = \frac{{b\tan {\theta _1} - b\tan {\theta _2}}}{{a\sec {\theta _1} - a\sec {\theta _2}}}\\\\ &\Rightarrow  \frac{{y - b\tan {\theta _1}}}{{x - a\sec {\theta _1}}} = \frac{b}{a}\left( {\frac{{\sin {\theta _1}\cos {\theta _2} - \cos {\theta _1}\sin {\theta _2}}}{{\cos {\theta _2} - \cos {\theta _1}}}} \right)\end{align}\]

Using simple trigonometric relations, this can be written as

\[\boxed{\frac{x}{a}\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) - \frac{y}{b}\sin \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right) = \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}\]

This is a useful general equation; it gives us the line joining any two points on the hyperbola. It would be advantageous to remember this equation.

Example - 8

When will a chord joining two points \({\theta _1}{\rm{\; and \;}}{\theta _2}\) on the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) pass through its focus (ae,0)?

Solution: We simply substitute \(x \to ae\) and \(y \to 0\) in the general equation for the chord we obtained above. Thus, we obtain

 \[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) = \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)\\ &\Rightarrow \quad\frac{{\cos \left( \begin{align}{\frac{{{\theta _1} - {\theta _2}}}{2}}\end{align} \right)}}{{\cos \left(\begin{align} {\frac{{{\theta _1} + {\theta _2}}}{2}}\end{align} \right)}} = \frac{1}{e}\end{align}\]

This is the required condition that must be satisfied. To write it more simply, we apply C and D to obtain.

\[\begin{align}&\frac{\begin{align}{\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) + \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}\end{align}}{\begin{align}{\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) - \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}\end{align}} = \frac{{1 + e}}{{1 - e}}\end{align}\]

This finally yields

\[\tan \left( {\frac{{{\theta _1}}}{2}} \right)\tan \left( {\frac{{{\theta _2}}}{2}} \right) = \frac{{1 - e}}{{1 + e}}\]

Example - 9

Let us define the exterior and interior regions of a hyperbola \(\begin{align}S(x,\,\,y):\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) as follows:

How can we decide whether a point \(P({x_1},{y_1})\) lies in the interior or exterior region of the hyperbola?

Solution: Let us consider \(P({x_1},{y_1})\) lying outside the hyperbola as shown below:

We have dropped a perpendicular from P onto the x-axis intersecting the hyperbola at\(Q({x_1},{y_0}).\) Since Q lies on the hyperbola, we have

\[\begin{align}&\!\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{x_1^2}}{{{a^2}}} - \frac{{y_0^2}}{{{b^2}}} = 1\\ &\Rightarrow \qquad y_0^2 = {b^2}\left( {\frac{{x_1^2}}{{{a^2}}} - 1} \right)\end{align}\]

Since \({y_1} > {y_0},\) we have

\[\begin{align}&\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y_1^2 > {b^2}\left( {\frac{{x_1^2}}{{{a^2}}} - 1} \right)\\ &\Rightarrow \qquad \frac{{x_1^2}}{{{a^2}}} - \frac{{y_1^2}}{{{b^2}}} - 1 < 0\\\\ &\Rightarrow \qquad S\left( {{x_1},\,\,{y_1}} \right) < 0\end{align}\]

Convince yourself that for any point \(P({x_1},\,\,{y_1})\) lying exterior to the hyperbola, this condition must be satisfied. Similarly for any point \(P({x_1},\,\,{y_1})\) lying in the interior region, we must have \(S({x_1},{y_1}) > 0.\)

We can summarize this discussion as:

\(S({x_1}\;,\;{y_1}) > 0\quad  \Rightarrow \quad P({x_1}\;,\;{y_1})\;\;\text{ lies inside the hyperbola.}\)

\(S({x_1}\;,\;{y_1}) = 0 \quad \Rightarrow \quad  P({x_1}\;,\;{y_1}) \;\;\text{lies on the hyperbola}\)

\(S({x_1}\;,\;{y_1}) < 0 \quad \Rightarrow \quad P({x_1}\;,\;{y_1}) \;\;\text{lies outside the hyperbola}\).

Example - 10

Find the condition that must be satisfied if the line \(y = mx + c\) is to intersect the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) in two distinct points. As a corollary, when will this line be a tangent to the hyperbola ?

Solution: The requisite condition can be obtained by solving simultaneously the equations of the line and the hyperbola.

\[\begin{align}&\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = mx + c\\ \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\qquad\quad\;\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\\ \\&\Rightarrow \qquad \frac{{{x^2}}}{{{a^2}}} - \frac{{{{(mx + c)}^2}}}{{{b^2}}} = 1\\ \\&\Rightarrow \qquad ({a^2}{m^2} - {b^2}){x^2} + 2{a^2}cmx + {a^2}({c^2} + {b^2}) = 0\end{align}\]

This quadratic will have two distinct roots if its discriminant is positive, i.e., if

\[\begin{align}&\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{a^4}{m^2}{c^2} > 4{a^2}({b^2} + {c^2})({a^2}{m^2} - {b^2})\\\\ &\Rightarrow \qquad {a^2}{m^2}{c^2} > {a^2}{m^2}{b^2} - {b^4} + {a^2}{m^2}{c^2} - {b^2}{c^2}\\\\ &\Rightarrow \qquad {c^2} > {a^2}{m^2} - {b^2}\end{align}\]

If this condition is satisfied, the line will intersect the hyperbola in two distinct points. It is clear that the line will be a tangent if

\[\begin{align}&\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c^2} = {a^2}{m^2} - {b^2}\\\\ &\Rightarrow \qquad {c^2} = \pm \sqrt {{a^2}{m^2} - {b^2}}\end{align}\]

Thus, we can conclude that any line of the form \(y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} \) will always be a tangent to the hyperbola \(\begin{align}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,\end{align}\) whatever be the value of m.

As an exercise prove that the coordinates of the point of contact of this tangent(s) will be

\[\begin{align}&\left( { \pm \frac{{{a^2}m}}{{\sqrt {{a^2}{m^2} - {b^2}} }},\,\, \pm \frac{{{b^2}}}{{\sqrt {{a^2}{m^2} - {b^2}} }}} \right)\end{align}\]

We discuss about tangents in more detail in the next section.

TRY YOURSELF - I

Q. 1 Find the equation of the hyperbola with the following parameters:

\(\text{Focus}\quad \quad \quad \quad \,\,\,\,:\) \((1,2)\)
\(\text{Directrix}\quad \quad \quad \,\,:\) \(2x + y - 1 = 0\)
\(\text{Eccentricity}\quad \quad \,:\) \(\sqrt 3 \)

Q. 2 Find the equation of the hyperbola with foci at \((0,\, \pm \sqrt {10} )\) and which passes through (2, 3).

Q. 3 Find the equation of the hyperbola whose conjugate axis is of length 5 and the distance between whose foci is 13.

Q. 4 Prove that the locus of the point of intersection of the lines

\[\begin{array}{l}\sqrt 3 x - y - 4\sqrt 3 \lambda = 0\\\sqrt 3 \lambda x + \lambda y - 4\sqrt 3 = 0\end{array}\]

for different values of \(\lambda \) is a hyperbola. What is the eccentricity of this hyperbola?

Q. 5 For what value of k does the line \(3x - y + k = 0\) touch the hyperbola \(\begin{align}\frac{{{x^2}}}{5} - \frac{{{y^2}}}{9} = 1\,\,?\end{align}\)