Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?

Question Image
Grade:11

1 Answers

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free