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?

Question

Samyak Jain · Answer

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 P as (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 = a 2 Pls approve.

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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?

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?

1 Answers

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