Consider the two curves $$ \begin{aligned} & C_{1}: y^{2}=4 x \\ & C_{2}: x^{2}+y^{2}-6 x+1=0 \end{aligned} $$ Then,

(B) $C_{1}$ and $C_{2}$ touch each other exactly at two points

(A) $C_{1}$ and $C_{2}$ touch each other only at one point

(C) $C_{1}$ and $C_{2}$ intersect (but do not touch) at exactly two points

(D) $C_{1}$ and $C_{2}$ neither intersect nor touch each other

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Parabola

Consider the two curves

C_{1} : y^{2} = 4 x C_{2} : x^{2} + y^{2} - 6 x + 1 = 0

Then,

Answer: (B)

The curve

C_{1} : y^{2} = 4 x

is a parabola with vertex at the origin and focus at

(1, 0)

. The curve

C_{2} : x^{2} + y^{2} - 6 x + 1 = 0

can be rewritten as

(x - 3)^{2} + y^{2} = 8

, which is a circle with center

(3, 1)

and radius

22

.

To find points of intersection, substitute

x = \frac{y ^{2}}{4}

from

C_{1}

into

C_{2}

\frac{y ^{4}}{16} + y^{2} - \frac{6 y ^{2}}{4} + 1 = 0

y^{4} + 16 y^{2} - 24 y^{2} + 16 = 0

y^{4} - 8 y^{2} + 16 = 0

(y^{2} - 4)^{2} = 0

This gives

y^{2} = 4

, so

y = \pm 2

, and correspondingly

x = 1

. The two curves touch at the points

(1, 2)

and

(1, - 2)

.

At these points, both curves have the same tangent line, confirming they touch rather than merely intersect.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option