UPSC NDA I 2021 Mathematics Solved Paper

CONCEPT:
The eccentricity of a hyperbola of the form ‌

y2

a2

−‌

x2

b2

=1‌‌ is given by: e=√1+‌

b2

a2

CALCULATION:
Given: The point (3‌tan‌θ,2‌sec‌θ ) lies on the hyperbola. As we can see that, if we substitutex=3‌tan‌θ and y=2‌sec‌θ in the expression
‌

y2

4

−‌

x2

9

we get ⇒‌

y2

4

−‌

x2

9

=sec2θ−tan2θ=1
So, we can say that, the point (3‌tan‌θ,2‌sec‌θ ) lies on the hyperbola ‌

y2

4

−‌

x2

9

=1
As we know that, eccentricity of the hyperbola of the form ‌

y2

a2

−‌

x2

b2

=1 is given by: e=√1+‌

b2

a2

Here, a2=4 and b2=9
⇒e=√1+‌

9

4

=‌

√13

2

Hence, correctoption is 4 .