# 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?

Samyak Jain
333 Points
4 years ago
Let coordinates of point P on the hyperbola be (a sec$\dpi{80} \large \Theta$, b tan$\dpi{80} \large \Theta$).
$\dpi{100} \therefore$ the coordinates of the foot of perpendicular N are (a sec$\dpi{80} \large \Theta$, 0).
Differentiate the equation of the hyperbola and replace a sec$\dpi{80} \large \Theta$ and b tan$\dpi{80} \large \Theta$ in place of x and y respectively.
You will get slope of tangent to the hyperbola at point P
as (b cosec$\dpi{80} \large \Theta$/ a).
Then equation of the tangent is y – b tan$\dpi{80} \large \Theta$ = (b cosec$\dpi{80} \large \Theta$/ a)(x – a sec$\dpi{80} \large \Theta$).
Put x = 0 in the above equation to get cooordinates of T(which is on x-axis).
$\dpi{100} \equiv$ (a cos$\dpi{80} \large \Theta$, 0)
Now, N(a sec$\dpi{80} \large \Theta$, 0), T(a cos$\dpi{80} \large \Theta$, 0), O(0,0),
by distance formula,
OT = $\dpi{100} \sqrt${(a cos$\dpi{80} \large \Theta$ – 0)– (0 – 0)2}  = a cos$\dpi{80} \large \Theta$             …. (1)
and ON = $\dpi{100} \sqrt${(a sec$\dpi{80} \large \Theta$ – 0)– (0 – 0)2}  = a sec$\dpi{80} \large \Theta$      …. (2)
From (1) & (2),
OT . ON  =  a cos$\dpi{80} \large \Theta$ . a sec$\dpi{80} \large \Theta$
$\dpi{100} \therefore$   OT . ON  =  a2
Pls approve.