Consider the two curves C_1: y^2=4x and C_2: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the points of intersection,substitute $y^2 = 4x$ into the equation of the circle $x^2 + y^2 - 6x + 1 = 0$.This gives $x^2 + 4x - 6x + 1 =

Vedclass · Accepted Answer

$C_1$ and $C_2$ touch each other exactly at two points.

Vedclass · Answer

(B) To find the points of intersection,substitute $y^2 = 4x$ into the equation of the circle $x^2 + y^2 - 6x + 1 = 0$.This gives $x^2 + 4x - 6x + 1 = 0$,which simplifies to $x^2 - 2x + 1 = 0$.This is $(x - 1)^2 = 0$,so $x = 1$.Substituting $x = 1$ into $y^2 = 4x$,we get $y^2 = 4$,which means $y = 2$ or $y = -2$.Thus,the curves intersect at the points $(1, 2)$ and $(1, -2)$.To check if they touch,we compare the slopes of the tangents at these points.For the parabola $y^2 = 4x$,differentiating gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,slope $m_1 = 1$. At $(1, -2)$,slope $m_1 = -1$.For the circle $x^2 + y^2 - 6x + 1 = 0$,differentiating gives $2x + 2y \frac{dy}{dx} - 6 = 0$,so $\frac{dy}{dx} = \frac{3 - x}{y}$. At $(1, 2)$,slope $m_2 = \frac{3 - 1}{2} = 1$. At $(1, -2)$,slope $m_2 = \frac{3 - 1}{-2} = -1$.Since the slopes are equal at both points,the curves touch each other at exactly two points.

Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

Similar Questions

If the point $(1, 4)$ lies inside the circle $x^2 + y^2 - 6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes,then the set of all possible values of $p$ is the interval

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$, then the maximum value of $(PR + PS)^2$ is:

The locus of the midpoints of the chords of the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ which subtend an angle of $\frac{\pi}{3}$ radians at its circumference is:

The slopes of the focal chords of the parabola $y^2=32x$,which are tangents to the circle $x^2+y^2=4$,are

$ABCD$ is a square with side length $a$. Taking $AB$ and $AD$ as the coordinate axes,find the equation of the circle passing through the vertices of the square.

Vedclass Test Series

Exam Paper Generator

Online Exam Module