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# Let we have two points as $P(a\sec \theta ,b\tan \theta )$ and $Q(a\sec \phi ,b\tan \phi )$, where $\theta + \phi = \dfrac{\pi }{2}$, be two points on hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. If (h,k) is the point of intersection of the normal at P &Q, then k is equal to_______A) $\left[ {\dfrac{{{a^2} + {b^2}}}{a}} \right]$B) $- \left[ {\dfrac{{{a^2} + {b^2}}}{a}} \right]$C) $\dfrac{{{a^2} + {b^2}}}{{ - b}}$D) $- \left[ {\dfrac{{{a^2} + {b^2}}}{b}} \right]$

Last updated date: 25th Jul 2024
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Hint: This question is based on the chapter conic sections, Hyperbola. The hyperbola with foci on the x- axis can be represented as $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. In a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, the length of latus rectum is given by $\dfrac{{2{b^{^2}}}}{a}$.

Complete step-by-step solution:
According to the question, firstly, we will obtain the slope of normal as $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ at $(a\sec \theta ,b\tan \theta )$.
Now, we will differentiate it with respect to $x$, and we get:
$\Rightarrow \dfrac{{2x}}{{{a^2}}} - \dfrac{{2y}}{{{b^2}}} \times \dfrac{{dy}}{{dx}} = 0$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{b^2}}}{{{a^2}}} \times \dfrac{x}{y}$
Slope for normal at point $(a\sec \theta ,b\tan \theta )$ will be:
$\Rightarrow - \dfrac{{{a^2}b\tan \theta }}{{{b^2}a\sec \theta }} = - \dfrac{a}{b}\sin \theta$
Therefore, equation of normal at $(a\sec \theta ,b\tan \theta )$ is:
$\Rightarrow y - b\tan \theta = - \dfrac{a}{b}\sin \theta \left( {x - a\sec \theta } \right)$
$\Rightarrow \left( {a\sin \theta } \right)x + by = \left( {{a^2} + {b^2}} \right)\tan \theta$
$\Rightarrow ax + by\cos ec\theta = \left( {{a^2} + {b^2}} \right)\sec \theta \,\,\,\,\,\,\,\, ----- equation\,1$
Similarly, the equation of normal to $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ at: $\left( {a\sec \phi ,b\tan \phi } \right)$ is
$ax+by\cos ec\varphi = \left( {{a^2} + {b^2}} \right){\sec \varphi}\,\,\,\,\,\,\,\,\, ----- equation\,2$
On subtracting the equation 2 from the equation 1, we get:
$\Rightarrow b\left( {\cos ec\theta - \cos ec\phi } \right)y = \left( {{a^2} + {b^2}} \right)\left( {\sec \theta - \sec \varphi } \right)$
By Calculating ‘y’, we get:
$\Rightarrow y = \dfrac{{{a^2} + {b^2}}}{b}.\dfrac{{\sec \theta - \sec \varphi }}{{\cos ec\theta - \cos ec\phi }}$
But we know that:
$\Rightarrow y = \dfrac{{{a^2} + {b^2}}}{b}.\dfrac{{\sec \theta - \sec \left( {\dfrac{\pi }{2} - \theta } \right)}}{{\cos ec\theta - \cos ec\left( {\dfrac{\pi }{2} - \theta } \right)}}$
But as per the question:
$\left[ {\because \phi + \theta = \dfrac{\pi }{2}} \right]$
So, we get:
$\dfrac{{\sec \theta - \cos ec\theta }}{{\sec \theta - \sec \theta }} = - 1$
Thus, we get that$y = - \left( {\dfrac{{{a^2} + {b^2}}}{b}} \right)$
That is $k = - \left( {\dfrac{{{a^2} + {b^2}}}{b}} \right)$
Therefore, we get the final result, and it is clear that Option D is the right option.

Note: Hyperbola is the difference of distances of a set of points in a plane from two fixed points is constant. Any points on a hyperbola should always be compared with the standard equation of a hyperbola.