The equation of the common tangent touching t | vedclass

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

Question

(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.This line touches the circle $(x - 3)^2 + y^2 = 9$ (with center $(3,

Vedclass · Accepted Answer

$\sqrt{3}y = x + 3$

Vedclass · Answer

(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.This line touches the circle $(x - 3)^2 + y^2 = 9$ (with center $(3, 0)$ and radius $3$) if the perpendicular distance from the center to the line equals the radius:$3 = \left| \frac{m(3) - 0 + \frac{1}{m}}{\sqrt{m^2 + 1}} \right|$Squaring both sides:$9(m^2 + 1) = (3m + \frac{1}{m})^2$$9m^2 + 9 = 9m^2 + 6 + \frac{1}{m^2}$$3 = \frac{1}{m^2} \implies m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$.Since the tangent is above the $x$-axis,we choose the positive slope $m = \frac{1}{\sqrt{3}}$.Substituting $m$ into the tangent equation:$y = \frac{1}{\sqrt{3}}x + \sqrt{3}$Multiplying by $\sqrt{3}$:$\sqrt{3}y = x + 3$.

The equation of the common tangent touching the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $x$-axis is

Similar Questions

Find the equation of the tangent to the parabola $x^2 = y$ at one of the endpoints of its latus rectum in the first quadrant.

Let the tangent and normal at any point $P(at^2, 2at)$, $(a > 0)$, on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P, T$ and $G$ is

The common tangent to the parabolas $y^2=32x$ and $x^2=256y$ is:

If the three normals drawn to the parabola $y^{2} = 2x$ pass through the point $(a, 0)$ where $a \neq 0$,then $a$ must be greater than:

The length of the latus rectum of the parabola $9x^2 - 6x + 36y + 19 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module