eliminate teta from the relations x=a sec teta and y = b tan teta?

Question

math in 11standard isc syllabus

Answer 1

here, x/a = sec teta _____ (1)
and, y/b = tan teta ______(2)
squaring and subtracting, we get
x^2/a^2 - y^2/b^2 = sec^2 teta - tan^2 teta = 1
or, x^2/a^2 - y^2/b^2 = 1

Answer 2

Let A = teta
x=a sec A, y=b tan A
tan^2 A + 1 = sec^2 A
(y/b)^2 + 1 = (x/a)^2

Answer 3

x = a sec teta
so sec teta = x/a ...1

y= b tan teta
so tam teta = y/b ....2

as sec^2 teta = tan^ 2 teta + 1

(x/a) ^2 = (y/b)^ 2 +1

or (x/a) ^2 -(y/b)^2 = 1

Answer 4

we know that cot teta = cos teta/sin teta & tan teta = sin teta/cos teta Now take the LHS of the problem: i.e. cot teta - tan teta / 1-2sin squared teta which can be written as {(cos teta / sin teta) - (sin teta / cos teta)} / {1-2sin squared teta} By simplifying the Numeratos & Dinominator, we get {(cos squared teta - sin squared teta)/sin teta cos teta} / cos2 teta [since 1-2sin ssquared teta = cos2 teta] = cos2 teta / (sin teta cos teta cos2 teta) = 1 / (sin teta cos teta) [cos2 teta get cancelled in Numerator & Dinominator] = (1/sin teta)(1/cos teta) = sec teta cosec teta [ since sec teta = 1/sin teta & cosec teta = 1/cos teta] which gives us the RHS of the problem. Hence it is proved that cot teta - tan teta / 1- 2sin squared teta = sec teta cosec teta

Answer 5

that damn question class XI th! im a class X th CBSE student.
divide x by a and y by b

x/a = sec θ and y/b = tan θ
we know (sec θ)^2 - (tan θ)^2=1
keeping values of x/a and y/b, we get

x^2/a^2 - y^2/b^2 = 1
that is the equation of a hyperbola.

Answer 6

Hint for your school homework problem:
Derive the equation of hyperbola in polar form to learn how to eliminate theta from the above question.

Answer 7

sec^2teta - tan^2 teta =1
therefore (x/a)^2 - (y/b)^2 = 1
(xb)^2 - (ya)^2 = (ab)^2

Answer 8

find x/a
find y/b
take squares of both sides
subtract the second from the first side by side
remember 1-sin^tetha=cos^2 tetha.
the result is a hyberbola equation.

Answer 9

(x^2/a^2)-(y^2/b^2)=1

Answer 10

sec theta = x/a
i.e. (sec theta )^2 = (x/a)^2
tan theta = y/b
i.e. (tan theta )^2 = (y/b)^2

But, (sec theta)^2 - (tan theta)^2 = 1

Hence, (x/a)^2 - (y/b)^2 = 1
It is the required eliminant.