TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Let P(h,k) be the pole of chord of the hyperbola
‌

x2

9

−‌

y2

36

=1 . . . (i)
Then, equation of chord
‌

hx

9

−‌

ky

36

=1. . . (ii)
Combining Eqs. (i) and (ii) and maintaining homogeneity
‌

x2

9

−‌

y2

36

=(‌

hx

9

−‌

ky

36

)2
∵ Chord substends right angle at the centre
⇒ Coefficient of x2+ Coefficient of y2=0
‌∴‌‌

1

9

−‌

h2

92

−‌

1

36

−‌

k2

362

‌‌=0
⇒‌‌

h2

92

+‌

k2

362

‌‌=‌

1

9

−‌

1

36

⇒‌16h2+k2‌‌=108
‌ Locus of pole ‌(h,k)
16x2+y2=108