The slope of the tangent to the circle (x-6)^ | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$.The eq

Vedclass · Accepted Answer

$\pm 1$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$.The equation of a line passing through $(4, 0)$ with slope $m$ is $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.This line is a tangent to the circle $(x-6)^2 + y^2 = 2$. The center of the circle is $(6, 0)$ and its radius is $r = \sqrt{2}$.The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$:$\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$Squaring both sides:$\frac{4m^2}{m^2 + 1} = 2$$4m^2 = 2m^2 + 2$$2m^2 = 2$$m^2 = 1$$m = \pm 1$.

The slope of the tangent to the circle $(x-6)^2 + y^2 = 2$,which passes through the focus of the parabola $y^2 = 16x$,is:

Similar Questions

The focal chord of the parabola $(y - 2)^2 = 16(x - 1)$ is a tangent to the circle $x^2 + y^2 - 14x - 4y + 51 = 0$. Then the slope of the focal chord can be:

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point,if ${k^2} =$

Let $A$ be the centre of the circle $x^{2}+y^{2}-2x-4y-20=0$. Let $B(1,7)$ and $D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is

Find the equation of the circle having normals $(x-1)(y-2)=0$ and a tangent $3x+4y=6$.

The combined equation of the tangent and normal to the curve $xy = 100$ at the point $(5, 20)$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module