Question

Tamgents are grom point on hyperbola x2/9+y2/4=1 to the circle x2+y2=9

Answer 1

Answer:

Let any point on the hyperbola be (3 sec θ, 2 tan θ).

Therefore, chord of contact of the circle x2 + y2 = 9 with respect to the point (3 sec θ, 2 tan θ) is,

(3 sec θ)x + (2 tan θ)y = 9.  …... (1)

Let (x1, y1) be the mid-point of the chord of contact.

Then the equation of the chord in mid point form is

xx1 + yy1 = x12 + y12  ….. (2)

Since both the above equations are identically equal, so

3 sec θ/x1 = 2 tan θ/y1 = 9 (x12 + y12)

This gives sec θ = 9x1/3(x12 + y12)

and tan θ = 9y1/2(x12 + y12)

Eliminating θ from the above two equations we get  

81x12/ 9(x12 + y12)2 – 81y12/4(x12 + y12)2 = 1.

Therefore, the required locus is x2/9 – y2/4 = (x2 + y2)2/ 81.

Answer 2

Step-by-step explanation:

The given equation of circle

is : x2+y2=9 ...(1)

and ellips is :x24+y28=1 ...(2)

From eqn. (1) and (2), we get

x24+9−x28=1

⇒ 2x2+9−x2=8⇒ x2=−1