Question

If the angle between asymptotes of the hyperbola \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[{{120}^{o}}\] and the product of the perpendiculars drawn from foci upon its any tangent is 9, then the locus of the point of intersection of the perpendicular tangents of the hyperbola can be
\[\left( \text{a} \right)\text{ }{{x}^{2}}+{{y}^{2}}=6\]
\[\left( \text{b} \right)\text{ }{{x}^{2}}+{{y}^{2}}=9\]
\[\left( \text{c} \right)\text{ }{{x}^{2}}+{{y}^{2}}=3\]
\[\left( \text{d} \right)\text{ }{{x}^{2}}+{{y}^{2}}=18\]

Answer 1

Hint: To solve the question, we will first find out the values of a and b. We will first find out the value of b or we will consider the tangent of the hyperbola as, \[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}.\] From this line, we will find the distance of both the foci and take their product as 9. This will give the value of b. After this, we will find the acute angle between the tangents and then we will apply the formula: y = mx + c to find ‘a’ in terms of ‘b’. Then we will find the intersection between the two tangents of a parabola having the slope m and \[\dfrac{-1}{m}\] to determine the focus.

Complete step by step solution:
In this question, our first aim is to find the values of a and b. For finding the value of b, we are given that the product of perpendicular distances from both the foci to any tangent is 9.

Here, the CD is the tangent, and SP and S’P’ are perpendiculars on this tangent where S (ae, 0) and S’ (– ae, 0 ) are the foci of the hyperbola. Here, e is the eccentricity of the hyperbola. We are given that,
\[SP\times {{S}^{'}}{{P}^{'}}=9......\left( i \right)\]
We will find the value of SP and S’P’ first. We know that if a point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given then its perpendicular distance from the line ax + by + c = 0 will be given by,
\[\text{distance }=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\]
The equation of the tangent CD is \[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}.\]
So, for SP, we have point \[\left( {{x}_{1}},{{y}_{1}} \right)\] as (ae, 0) and the line ax + by + c = 0 as \[mx-y+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}.\] Therefore, we get the distance SP as,
\[SP=\left| \dfrac{m\left( ae \right)-0+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}}{\sqrt{{{m}^{2}}+1}} \right|.....\left( ii \right)\]
So, for S’P’, we have point \[\left( {{x}_{1}},{{y}_{1}} \right)\] as (– ae, 0) and the line ax + by + c = 0 as \[mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}.\] Therefore, we get the distance S’P’ as,
\[{{S}^{'}}{{P}^{'}}=\left| \dfrac{m\left( -ae \right)-0+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}}{\sqrt{{{m}^{2}}+1}} \right|.....\left( iii \right)\]
Now, we will put the values of SP and S’P’ in equation (i).
\[\Rightarrow \left| \dfrac{m\left( ae \right)-0+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}}{\sqrt{{{m}^{2}}+1}} \right|\left| \dfrac{m\left( -ae \right)+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}}{\sqrt{{{m}^{2}}+1}} \right|=9\]
\[\Rightarrow \left| \dfrac{\left( ame+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}} \right)\left( \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}-ame \right)}{{{m}^{2}}+1} \right|=9\]
In the above equation, we will use the identity: \[\left( {{a}^{'}}+{{b}^{'}} \right)\left( {{a}^{'}}-{{b}^{'}} \right)={{a}^{2'}}-{{b}^{2'}}.\] Here, \[{{a}^{'}}=\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\] and b’ = ame. Thus, we will get,
\[\Rightarrow \left| \dfrac{{{\left( \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}} \right)}^{2}}{{\left( ame \right)}^{2}}}{{{m}^{2}}+1} \right|=9\]
\[\Rightarrow \left| \dfrac{{{a}^{2}}{{m}^{2}}-{{b}^{2}}-{{a}^{2}}{{m}^{2}}{{e}^{2}}}{{{m}^{2}}+1} \right|=9......\left( iv \right)\]
Now we know that e is the eccentricity and its value is given by,
\[e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}.....\left( v \right)\]
Now, we will put the value of e from (v) to (iv). Thus, we will get,
\[\Rightarrow \left| \dfrac{{{a}^{2}}{{m}^{2}}-{{b}^{2}}-{{a}^{2}}{{m}^{2}}\left( 1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)}{{{m}^{2}}+1} \right|=9\]
\[\Rightarrow \left| \dfrac{{{a}^{2}}{{m}^{2}}-{{b}^{2}}-{{a}^{2}}{{m}^{2}}-{{m}^{2}}{{b}^{2}}}{{{m}^{2}}+1} \right|=9\]
\[\Rightarrow \left| \dfrac{-{{b}^{2}}\left( {{m}^{2}}+1 \right)}{{{m}^{2}}+1} \right|=9\]
\[\Rightarrow \left| -{{b}^{2}} \right|=9\]
\[\Rightarrow {{b}^{2}}=9.....\left( vi \right)\]
Now, we are given that, \[\angle BO{{A}^{'}}={{120}^{o}}.\] Therefore, we have,
\[\angle BO{{A}^{'}}+\angle {{A}^{'}}OA={{180}^{o}}\]
\[\Rightarrow {{120}^{o}}+\angle {{A}^{'}}OA={{180}^{o}}\]
\[\Rightarrow \angle {{A}^{'}}OA={{60}^{o}}\]
Thus,
\[\angle {{A}^{'}}OX=\dfrac{1}{2}\angle {{A}^{'}}OA=\dfrac{1}{2}\times {{60}^{o}}\]
\[\Rightarrow \angle {{A}^{'}}OX={{30}^{o}}\]
\[\Rightarrow x={{30}^{o}}\]
The equation of the asymptote A’B’ is given by
\[y=\dfrac{b}{a}x....\left( vii \right)\]
Now, the equation of A’B’ can also be written as a slope-intercept form. Thus,
\[y=mx+c\]
\[\Rightarrow y=\left( \tan {{30}^{o}} \right)x+0\]
\[\Rightarrow y=\dfrac{x}{\sqrt{3}}....\left( viii \right)\]
From (vii) and (viii),
\[\dfrac{b}{a}=\dfrac{1}{\sqrt{3}}\]
On squaring both the sides, we get,
\[\Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{1}{3}\]
\[\Rightarrow {{a}^{2}}=3{{b}^{2}}....\left( ix \right)\]
From (ix) and (vi), we have,
\[\Rightarrow {{a}^{2}}=3\times 9\]
\[\Rightarrow {{a}^{2}}=27\]
Now, the equation of hyperbola becomes,
\[\dfrac{{{x}^{2}}}{27}-\dfrac{{{y}^{2}}}{9}=1\]
Now, the equation of tangent will be,
\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\]
\[\Rightarrow y=mx+\sqrt{27{{m}^{2}}-9}.....\left( x \right)\]
Now, there is another tangent having the slope \[{{m}_{1}}.\] Its equation is,
\[\Rightarrow y={{m}_{1}}x+\sqrt{27{{m}_{1}}^{2}-9}.....\left( xi \right)\]
Now, both the tangents are perpendicular to each other. So,
\[m{{m}_{1}}=-1\]
\[\Rightarrow {{m}_{1}}=\dfrac{-1}{m}.....\left( xii \right)\]
From (xi) and (xii), we have,
\[y=\dfrac{-1}{m}x+\sqrt{\dfrac{27}{{{m}^{2}}}-9}\]
\[\Rightarrow y=\dfrac{-x}{m}+\sqrt{\dfrac{27-9{{m}^{2}}}{{{m}^{2}}}}\]
\[\Rightarrow y=\dfrac{-x}{m}+\dfrac{\sqrt{27-9{{m}^{2}}}}{m}\]
\[\Rightarrow my=-x+\sqrt{27-9{{m}^{2}}}.....\left( xiii \right)\]
From (x), we have,
\[y=mx+\sqrt{27{{m}^{2}}-9}\]
\[\Rightarrow y-mx=\sqrt{27{{m}^{2}}-9}\]
On squaring both the sides, we get,
\[\Rightarrow {{\left( y-mx \right)}^{2}}={{\left( \sqrt{27{{m}^{2}}-9} \right)}^{2}}\]
Here, we will use the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.\] So, we will get,
\[{{y}^{2}}+{{m}^{2}}{{x}^{2}}-2mxy=27{{m}^{2}}-9......\left( xiv \right)\]
From (xiii), we have,
\[my+x=\sqrt{27-9{{m}^{2}}}\]
On squaring both the sides, we get,
\[{{\left( my+x \right)}^{2}}={{\left( \sqrt{27-9{{m}^{2}}} \right)}^{2}}\]
\[\Rightarrow {{m}^{2}}{{y}^{2}}+{{x}^{2}}+2mxy=27-9{{m}^{2}}.....\left( xv \right)\]
Now, we will add the equations (xiv) and (xv). Thus, we will get,
\[\left( {{y}^{2}}+{{m}^{2}}{{x}^{2}}-2mxy \right)+\left( {{m}^{2}}{{y}^{2}}+{{x}^{2}}+2mxy \right)=\left( 27{{m}^{2}}-9 \right)+\left( 27-9{{m}^{2}} \right)\]
\[\Rightarrow {{y}^{2}}+{{m}^{2}}{{x}^{2}}-2mxy+{{m}^{2}}{{y}^{2}}+{{x}^{2}}+2mxy=27{{m}^{2}}-9{{m}^{2}}-9+27\]
\[\Rightarrow {{y}^{2}}\left( {{m}^{2}}+1 \right)+{{x}^{2}}\left( {{m}^{2}}+1 \right)=18+18{{m}^{2}}\]
\[\Rightarrow {{y}^{2}}\left( {{m}^{2}}+1 \right)+{{x}^{2}}\left( {{m}^{2}}+1 \right)=18\left( {{m}^{2}}+1 \right)\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=18\]
The above equation is our required locus.
Hence, option (d) is the right answer.

 Note: After finding the values of ‘a’ and ‘b’, we can also approach the solution in the following way:
The intersections of perpendicular tangents of the hyperbola lie on the director circle of the hyperbola. Thus, the required locus is the director circle of the hyperbola obtained. The director circle of \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] is:
\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}-{{b}^{2}}\]
Hence, the director circle of \[\dfrac{{{x}^{2}}}{27}-\dfrac{{{y}^{2}}}{9}=1\] will be:
\[{{x}^{2}}+{{y}^{2}}=27-9\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=18\]
This is our required locus.