TS EAMCET 7 May 2018 Shift 1 Solved Paper

x2

4

+

y2

9

= 1 and

x2

16

−

y2

k

= 1
Solve these equations.
x2=

144+16k

36+k

,y2=

−27k

36+k

.....(I)
Differentiate

x2

4

+

y2

9

=1.

2x

4

+

2yy′

9

=0
y′=−

9

4

x

y

For the second hyperbola
y′=

k

16

x

y

since both the hyperbola's are orthogonal.
−

9

4

x

y

×

k

16

x

y

=−1
9kx2=64y2
Substitute this in equation (I).
9k(

144+16k

36+k

)=64(

−27

36+k

)
k=−21