Maths-

General

Easy

Question

# Consider the two curves C_{1} : y^{2} = 4xC_{2} : x^{2} + y^{2} – 6x + 1 = 0Then,

- C
_{1} and C_{2} touch each other only at one point
- C
_{1} and C_{2} touch each other exactly at two points
- C
_{1} and C_{2} intersect (but do not touch) at exactly two points
- C
_{1} and C_{2} neither intersect nor touch each other

_{1}and C_{2}touch each other only at one point_{1}and C_{2}touch each other exactly at two points_{1}and C_{2}intersect (but do not touch) at exactly two points_{1}and C_{2}neither intersect nor touch each other## The correct answer is: C_{1} and C_{2} touch each other exactly at two points

### To find the correct option.

Solving the two equations, we get $x=1$ and $y=±2$

So the two curves meet at two points $(1,2)$ and $(1,−2).$

Equation of the tangent at $(1,2)$ to $C_{1}$ is $y(2)=2(x+1)$

$y=x+1$

and Equation of the tangent at $(1,2)$ to $C_{2}$ is

$x.1+y(2)−3(x+1)+1=0$

y=x+1

Showing that $C_{1}$ and $C_{2}$ have a common tangent at the point $(1,2).$

Similarly they have a common tangent

$y=−(x+1)at(1,−2)$

So the two curves meet at two points $(1,2)$ and $(1,−2).$

Equation of the tangent at $(1,2)$ to $C_{1}$ is $y(2)=2(x+1)$

$y=x+1$

and Equation of the tangent at $(1,2)$ to $C_{2}$ is

$x.1+y(2)−3(x+1)+1=0$

Similarly they have a common tangent

$y=−(x+1)at(1,−2)$

Hence, the two curves touch each other exactly at two points.

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