Question

The common tangents to the circle x2+y2=2 and the parabola y2=8x

Answer 1

Answer:

Lets assume common tangents touches circle at P(x

1

,y

1

) and R(x

2

,y

2

)

Then,

C

1

(x

1

,y

1

):x

1

2

+y

1

2

−2=0

C

2

(x

2

,y

2

):y

2

2

−8x

2

=0

The tangent space to the conics is

m

1

=

dx

1

dy

1

=−

y

1

x

1

… (i)

m

2

=

dx

2

dy

2

=

y

2

4

…(ii)

The tangency condition reads:

{

y

2

−y

1

=m

1

(x

2

−x

1

) …(iii)

y

1

−y

2

=m

2

(x

1

−x

2

) …(iv)

On solving these four eqs, we get

x

1

=−1,x

2

=2,y

1

=±1,y

2

=±4

Therefore,

P≡(−1,1)

Q≡(−1,−1)

R≡(2,4)

S≡(2,−4)

Now the area of the quadrilateral PQRS:

2

1

×(PQ+RS)×(Distance between PQ and RS)

=

2

1

×(2+8)×(1+2)

=

2

1

×10×3=15 sq units.