Question

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola \[\dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{5} = 1\], meets x - axis and y - axis at A and B respectively. Then \[{\left( {OA} \right)^2} - {\left( {OB} \right)^2}\], where O is origin equals:
A) \[\dfrac{{ - 20}}{9}\]
B) \[\dfrac{{16}}{9}\]
C) \[4\]
D) \[\dfrac{4}{3}\]

Answer 1

Hint: Here first we will find the values of a and b by comparing the given equation of hyperbola with its standard equation i.e. \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] and then we will find the value of eccentricity which is given by:-
\[{b^2} = {a^2}\left( {{e^2} - 1} \right)\] now since the extremity of latus rectum is \[\left( {ae,\dfrac{{{b^2}}}{a}} \right)\] hence we will find the latus rectum and then we will find the equation of tangent passing through the latus rectum and then we will find the points A and B by putting x = 0 and y = 0 in the equation of tangent and then find their distance from origin and the compute the required value.

Complete step-by-step answer:
               

The standard equation of hyperbola is given by:-
\[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]
Comparing the given equation of hyperbola with standard equation we get:-
\[{a^2} = 4\] \[\&\] \[{b^2} = 5\]……………………..(1)
\[ \Rightarrow a = 2\] \[\&\] \[b = \sqrt 5 \]……………………….(2)
Now eccentricity for standard hyperbola is given by:-
\[{b^2} = {a^2}\left( {{e^2} - 1} \right)\]
Therefore, the eccentricity of the given hyperbola is:-
Putting values from equation 1 we get:-
\[5 = 4\left( {{e^2} - 1} \right)\]
Calculating the value of e we get:-
\[
  {e^2} = \dfrac{5}{4} + 1 \\
   \Rightarrow {e^2} = \dfrac{{5 + 4}}{4} \\
   \Rightarrow {e^2} = \dfrac{9}{4} \\
 \]
Taking square root of both sides we get:-
\[
  \sqrt {{e^2}} = \sqrt {\dfrac{9}{4}} \\
  e = \dfrac{3}{2} \\
 \]
Now since the latus rectum in 1st quadrant is given by:-
\[\left( {ae,\dfrac{{{b^2}}}{a}} \right)\]
Hence putting in the respective values from equation 1 and 2 we get:-
\[{\text{latus rectum}} \equiv \left( {2 \times \dfrac{3}{2},\dfrac{5}{2}} \right)\]
Simplifying it we get:-
\[{\text{latus rectum}} \equiv \left( {3,\dfrac{5}{2}} \right)\]
Now we will find the equation of tangent of the given hyperbola.
The standard equation of tangent of hyperbola passing through \[\left( {{x_1},{y_1}} \right)\] is given by:-
\[\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1\]
Now since it is given that the tangent passes through the latus rectum hence its equation is given by:-
\[\dfrac{{x\left( 3 \right)}}{4} - \dfrac{{y\left( {\dfrac{5}{2}} \right)}}{5} = 1\]
Simplifying it further we get:-
\[
  \dfrac{{x\left( 3 \right)}}{4} - \dfrac{{5y}}{{2 \times 5}} = 1 \\
   \Rightarrow \dfrac{{3x}}{4} - \dfrac{y}{2} = 1 \\
 \]
Now it is given that the tangent meets x axis at A hence y=0
Putting y=0 in above equation we get:-
\[\dfrac{{3x}}{4} - \dfrac{0}{2} = 1\]
Solving for x we get:-
\[
  \dfrac{{3x}}{4} = 1 \\
  x = \dfrac{4}{3} \\
 \]
Hence point A is \[A \equiv \left( {\dfrac{4}{3},0} \right)\]
Now it is given that the tangent meets y axis at B hence x=0
Putting x=0 in above equation we get:-
\[\dfrac{{3\left( 0 \right)}}{4} - \dfrac{y}{2} = 1\]
Solving for y we get:-
\[
  \dfrac{{ - y}}{2} = 1 \\
  y = - 2 \\
 \]
Hence point B is \[B \equiv \left( {0, - 2} \right)\]
Now we will find the distance of Point A from origin (0, 0).
The distance between two points \[\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)\] is given by:-
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Putting the values for point A and origin we get:-
\[OA = \sqrt {{{\left( {\dfrac{4}{3} - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} \]
Simplifying it further we get:-
\[
  OA = \sqrt {{{\left( {\dfrac{4}{3}} \right)}^2}} \\
   \Rightarrow OA = \dfrac{4}{3} \\
 \]
Now we will find the distance of Point B from origin (0, 0).
The distance between two points \[\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)\] is given by:-
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Putting the values for point A and origin we get:-
\[OB = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 2 - 0} \right)}^2}} \]
Simplifying it further we get:-
\[
  OB = \sqrt {{{\left( { - 2} \right)}^2}} \\
   \Rightarrow OB = \sqrt 4 \\
   \Rightarrow OB = 2 \\
 \]
Now we will evaluate the value of \[{\left( {OA} \right)^2} - {\left( {OB} \right)^2}\]
Putting in the respective values we get:-
\[{\left( {OA} \right)^2} - {\left( {OB} \right)^2} = {\left( {\dfrac{4}{3}} \right)^2} - {\left( 2 \right)^2}\]
Simplifying it further we get:-
\[{\left( {OA} \right)^2} - {\left( {OB} \right)^2} = \dfrac{{16}}{9} - 4\]
Taking the LCM we get:-
\[{\left( {OA} \right)^2} - {\left( {OB} \right)^2} = \dfrac{{16 - 4\left( 9 \right)}}{9}\]
Solving it further we get:-
\[
   \Rightarrow {\left( {OA} \right)^2} - {\left( {OB} \right)^2} = \dfrac{{16 - 36}}{9} \\
   \Rightarrow {\left( {OA} \right)^2} - {\left( {OB} \right)^2} = \dfrac{{ - 20}}{9} \\
 \]

So, the correct answer is “Option A”.

Note: Students should take a note that the coordinate of latus rectum in first quadrant is given by:-
\[\left( {ae,\dfrac{{{b^2}}}{a}} \right)\]
Also, all the points and calculations should be evaluated carefully to get the correct answer.