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?

Paridhi sharma Grade: 11

        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

Answers : (1)

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Samyak Jain

331 Points

							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  =  a²
Pls approve.