Question

If the tangent drawn to the hyperbola \[4{{y}^{2}}={{x}^{2}}+1\] intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid-point then AB is:
A. \[{{x}^{2}}-4{{y}^{2}}+16{{x}^{2}}{{y}^{2}}=0\]
B. \[4{{x}^{2}}-{{y}^{2}}+16{{x}^{2}}{{y}^{2}}=0\]
C. \[4{{x}^{2}}-{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\]
D. \[{{x}^{2}}-4{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\]

Answer 1

Hint: We need to rewrite the given equation in the hyperbola form of equation and find out the value of ‘a’ and ‘b’. Then we will find out the value of the slope by differentiating the given equation of hyperbola as \[\dfrac{dy}{dx}\] is the value of the slope. Then let the equation of the common tangent be \[y=mx+c\] and substituting the value of ‘m’ in this and simplify the following. Thus, we get the coordinates of the point when it intersects x-axis and at the point intersects y-axis. Then finding the value of ‘h’ and ‘k’ and substituting it in the given equation, we will get the locus of the mid-point of AB.

Formula used:
Equation of the parabola is of the form;
\[ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
The equation of the common tangent is of the form \[y=mx+c\] .

Complete step by step solution:
We have given that,
The tangent drawn to the hyperbola \[4{{y}^{2}}={{x}^{2}}+1\] intersects the co-ordinate axes at the distinct points A and B.
We have,
\[\Rightarrow 4{{y}^{2}}={{x}^{2}}+1\]
\[\Rightarrow -{{x}^{2}}+4{{y}^{2}}=1\]
Rewrite in the form of equation of hyperbola;
\[\Rightarrow -\dfrac{{{x}^{2}}}{{{1}^{2}}}+\dfrac{4{{y}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}}=1\]
Thus,
\[a=1\ and\ b=\dfrac{1}{2}\]
Now,
Let the tangent to the curve is at the point \[\left( {{x}_{1}},{{y}_{1}} \right)\]
Thus,
\[\Rightarrow 4{{y}_{1}}^{2}={{x}_{1}}^{2}+1\]
Differentiate it with respect to ‘x’,
\[\Rightarrow 4\times 2{{y}_{1}}\cdot \dfrac{dy}{dx}=2{{x}_{1}}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{8y}=\dfrac{{{x}_{1}}}{4{{y}_{1}}}\]
So,
The slope = \[m=\dfrac{{{x}_{1}}}{4{{y}_{1}}}\]
Let the equation of the common tangent be \[y=mx+c\]
Substituting the value of ‘m’,
\[\Rightarrow y=mx+c\]
\[\Rightarrow {{y}_{1}}=\left( \dfrac{{{x}_{1}}}{4{{y}_{1}}} \right)x+c=\dfrac{{{x}_{1}}^{2}}{4{{y}_{1}}}+c\]
Thus, solving for the value of ‘c’,
\[\Rightarrow c={{y}_{1}}-\dfrac{{{x}_{1}}^{2}}{4{{y}_{1}}}=\dfrac{4{{y}_{1}}^{2}-{{x}_{1}}^{2}}{4{{y}_{1}}}\]
\[\Rightarrow \dfrac{4{{y}_{1}}^{2}-{{x}_{1}}^{2}}{4{{y}_{1}}}=\dfrac{1}{4{{y}_{1}}}\]
Thus,
The value of ‘y’,
\[\Rightarrow y=\dfrac{{{x}_{1}}}{4{{y}_{1}}}x+\dfrac{1}{4{{y}_{1}}}\]
\[\Rightarrow 4{{y}_{1}}y={{x}_{1}}x\] ------ (1)
Hence,
We got the coordinates of the x-axis and the y-axis;
Intersects x-axis at \[\left( \dfrac{-1}{{{x}_{1}}},0 \right)\] and intersect at y-axis at \[\left( 0,\dfrac{1}{4{{y}_{1}}} \right)\]
Let the mid-point ‘M’ \[\left( h,k \right)\] of AB,
Thus,
\[\Rightarrow 2h=-\dfrac{1}{{{x}_{1}}}\Rightarrow {{x}_{1}}=-\dfrac{1}{2h}\]
And
\[\Rightarrow 2k=\dfrac{1}{4{{y}_{1}}}\Rightarrow {{y}_{1}}=\dfrac{1}{8k}\]
Since the point \[P\left( {{x}_{1}},{{y}_{1}} \right)\] lies on the hyperbola so it will satisfy this point.
We have,
\[\Rightarrow 4{{y}_{1}}^{2}={{x}_{1}}^{2}+1\]
Substituting the values,
\[\Rightarrow 4{{\left( \dfrac{1}{8k} \right)}^{2}}={{\left( -\dfrac{1}{2h} \right)}^{2}}+1\]
Solving the above we will get
\[\Rightarrow 4\left( \dfrac{1}{64{{k}^{2}}} \right)=\left( -\dfrac{1}{4{{h}^{2}}} \right)+1\]
Cancelling out the common terms, we will get
\[\Rightarrow \dfrac{1}{16{{k}^{2}}}=\left( -\dfrac{1}{4{{h}^{2}}} \right)+1\]
Simplifying the above, we will get
\[\Rightarrow 4{{h}^{2}}=16{{k}^{2}}\left( 1+4{{h}^{2}} \right)\]
It can written as,
\[\Rightarrow {{x}^{2}}=4{{y}^{2}}+16{{x}^{2}}{{y}^{2}}\]
Or
\[\Rightarrow {{x}^{2}}-4{{y}^{2}}-16{{x}^{2}}{{y}^{2}}=0\]

Hence, the option (D) is the correct answer.

Note: Students should need to remember that or should note that the equation of the locus with the given mid-point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given by;
\[\dfrac{x{{x}_{1}}}{{{a}^{2}}}-\dfrac{y{{y}_{1}}}{{{b}^{2}}}=\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}-\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}\]
In this type of questions where it is asked to find the locus of something, follow all the given conditions accordingly to the questions then we will get the required equation of the locus.