Question

Let P(3, 3) be a point on the hyperbola, \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] . If the normal to it at P intersects the x axis at (9, 0) and e is the eccentricity, then the ordered pair \[({a^2},{e^2})\] is equal to
A. (9, 3)
B. \[\left( {\dfrac{9}{2},2} \right)\]
C. \[\left( {\dfrac{9}{2},3} \right)\]
D. \[\left( {\dfrac{3}{2},2} \right)\]

Answer 1

Hint: Here we have to find the ordered pair \[({a^2},{e^2})\]. As they given in the question the equation of hyperbola i.e., \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] at the point P(3, 3) and the equation of the normal intersecting at the point (9, 0). By using these two equations we can determine the value of \[{a^2}\] and \[{e^2}\].

Formula Used:
The standard equation of hyperbola is \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\].
The equation of normal to the given hyperbola at the point \[P({x_1},{y_1})\] is given by \[\dfrac{{ax}}{{{x_1}}} + \dfrac{{by}}{{{y_1}}} = {a^2}{e^2}\]

Complete step by step Solution:
The equation of hyperbola is given by \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], then the equation at the point P (3, 3), i.e., value of x is 3 and the value of y is 3. On substituting the values of x and y in the hyperbola equation we get
\[\dfrac{9}{{{a^2}}} - \dfrac{9}{{{b^2}}} = 1\]……….. (i)
The equation of normal to the given hyperbola at the point \[P({x_1},{y_1})\] is given by \[\dfrac{{ax}}{{{x_1}}} + \dfrac{{by}}{{{y_1}}} = {a^2}{e^2}\]
Here the value of \[{x_1}\] is 3 and the value of \[{y_1}\] is 3. Then the value of x is 9 and the value of y is 0. On substituting these values in the equation of normal we get
\[\dfrac{{{a^2}9}}{3} + \dfrac{{{b^2}(0)}}{3} = {a^2}{e^2}\]
On simplifying we get
\[{e^2} = 3\]
As we know that the eccentricity formula for the hyperbola is given by \[{e^2} = 1 + \dfrac{{{b^2}}}{{{a^2}}}\]
On substituting the value of eccentricity in the above formula
\[3 = 1 + \dfrac{{{b^2}}}{{{a^2}}}\]
On simplifying it we get
\[{b^2} = 2{a^2}\]……… (ii)
On substituting the equation (ii) in the equation (i)
\[\dfrac{9}{{{a^2}}} - \dfrac{9}{{2{a^2}}} = 1\]
On taking LCM we get
\[\dfrac{{18 - 9}}{{2{a^2}}} = 1\]
On simplifying we get
\[\dfrac{9}{{2{a^2}}} = 1\]
\[\dfrac{9}{2} = {a^2}\]
Therefore the ordered pair is \[\left( {\dfrac{9}{2},3} \right)\]

Hence, the correct option is C.

Note: The formula for the eccentricity of any conic section is defined as \[e = \dfrac{c}{a}\]. Where c is the distance from the centre to the focus and a is the distance from the centre to the vertex. Remember the eccentricity for circle is 0, the eccentricity for parabola is 1 and the eccentricity for the ellipse is \[\sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \]