# 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) \[ - \left[ {\dfrac{{{a^2} + {b^2}}}{b}} \right]\]

Answer

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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.

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