Question

A hyperbola passes through the point $ P(\sqrt 2 ,\sqrt 3 ) $ and has foci at $ ( \pm 2,0) $ . Then the tangent to this hyperbola at P also passes through the point.
A. $ (3\sqrt 2 ,2\sqrt 3 ) $
B. $ (2\sqrt 2 ,3\sqrt 3 ) $
C. $ (\sqrt 3 ,\sqrt 2 ) $
D. $ ( - \sqrt 2 , - \sqrt 3 ) $

Answer 1

Hint: A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone. The hyperbola is a set of all the points that have the same difference between the distances to each focus. Each hyperbola has two foci. Then use the equation of the tangent to find out the other point through which the tangent passes.

Complete step-by-step answer:
The given hyperbola has foci $ ( \pm 2,0) $ that lie on the x-axis.
Equation of a hyperbola with horizontal transverse is given as,
 $ \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 $
We know the coordinates of foci are $ ( \pm 2,0) $ that is
 $ ae = 2 $
Squaring both sides we get $ {a^2}{e^2} = 4 $
We know,
 $
  {a^2} + {b^2} = {a^2}{e^2} \\
   \Rightarrow {a^2} + {b^2} = 4 \\
  {b^2} = 4 - {a^2} \;
  $
Put this value in the equation of hyperbola,
 $ \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{4 - {a^2}}} = 1 $
We are given that the hyperbola passes through the point $ P(\sqrt 2 ,\sqrt 3 ) $ , so we get –
 $
  \dfrac{{{{(\sqrt 2 )}^2}}}{{{a^2}}} - \dfrac{{{{(\sqrt 3 )}^2}}}{{4 - {a^2}}} = 1 \\
  \dfrac{2}{{{a^2}}} - \dfrac{3}{{4 - {a^2}}} = 1 \\
  \dfrac{{2(4 - {a^2}) - 3{a^2}}}{{{a^2}(4 - {a^2})}} = 1 \\
  8 - 2{a^2} - 3{a^2} = 4{a^2} - {a^4} \;
  $
On rearranging the above equation, we get a polynomial –
 $
  {a^4} - 5{a^2} - 4{a^2} + 8 = 0 \\
  {a^4} - 9{a^2} + 8 = 0 \\
  $
Let us find the solution of the above equation,
\[
  {a^4} - {a^2} - 8{a^2} + 8 = 0 \\
  {a^2}({a^2} - 1) - 8({a^2} - 1) = 0 \\
  ({a^2} - 8)({a^2} - 1) = 0 \\
   \Rightarrow {a^2} = 8,{a^2} = 1 \\
   \Rightarrow a = \pm 2\sqrt 2 ,a = \pm 1 \;
 \]
Now, we know that
 $
  ae = 1 \\
   \Rightarrow e = \dfrac{1}{a}, \\
  e \geqslant 1 \;
  $
Thus the negative value of $ a $ cannot be possible.
At $ a = 2\sqrt 2 ,e = \dfrac{1}{{2\sqrt 2 }} $ which is smaller than 1, so $ a $ cannot be equal to $ 2\sqrt 2 $ .
Thus only possible value is $ a = 1 $ .
Now, we know
 $
  {b^2} = 4 - {a^2} \\
   \Rightarrow {b^2} = 4 - 1 = 3 \;
  $
The equation of the hyperbola becomes,
 $ \dfrac{{{x^2}}}{1} - \dfrac{{{y^2}}}{3} = 1 $
The equation of the tangent at point P is given as,
 $
  \dfrac{{\sqrt 2 x}}{1} - \dfrac{{\sqrt 3 y}}{3} = 1 \\
  \sqrt 2 x - \dfrac{{\sqrt 3 y}}{3} = 1 \;
  $
Now, let us check which option is correct by putting their values in the above equation.
A. $ (3\sqrt 2 ,2\sqrt 3 ) $
Put $ x = 3\sqrt 2 $ and $ y = 2\sqrt 3 $ in the equation of tangent.
 $ \sqrt 2 \times 3\sqrt 2 - \dfrac{{\sqrt 3 \times 2\sqrt 3 }}{3} = 6 - 2 = 4 \ne 1 $
Thus, $ (3\sqrt 2 ,2\sqrt 3 ) $ is not the right answer.
B. $ (2\sqrt 2 ,3\sqrt 3 ) $
Put $ x = 2\sqrt 2 $ and $ y = 3\sqrt 3 $ in the equation of tangent.
 $ \sqrt 2 \times 2\sqrt 2 - \dfrac{{\sqrt 3 \times 3\sqrt 3 }}{3} = 4 - 3 = 1 $
These coordinates satisfy the equation of tangent.
So, the correct answer is “Option B”.

Note: The graph of a hyperbola has two distinct branches, thus the curve is not continuous. The line segment containing both the foci of the hyperbola whose endpoints are both on the hyperbola is called the transverse axis. The midpoint of the transverse axis (the point halfway between foci) is called the centre.