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        I want to request the solution for question no. 12. I tried solving it by finding the equation of normal at parametric point P and putting y= 0 but didn't get the answer. What's wrong with it?
one year ago

							Let coordinates of point P on the hyperbola be (a sec, b tan). the coordinates of the foot of perpendicular N are (a sec, 0).Differentiate the equation of the hyperbola and replace a sec and b tan in place of x and y respectively.You will get slope of tangent to the hyperbola at point Pas (b cosec/ a).Then equation of the tangent is y – b tan = (b cosec/ a)(x – a sec).Put x = 0 in the above equation to get cooordinates of T(which is on x-axis).T  (a cos, 0)Now, N(a sec, 0), T(a cos, 0), O(0,0),by distance formula,OT = {(a cos – 0)2 – (0 – 0)2}  = a cos             …. (1)and ON = {(a sec – 0)2 – (0 – 0)2}  = a sec      …. (2)                       From (1) & (2),OT . ON  =  a cos . a sec     OT . ON  =  a2Pls approve.

8 months ago
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