The focal chord to y^2 = 16x is tangent to (x | vedclass

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

Question

(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(4, 0)$.Any focal chord passing through $(4, 0)$ with

Vedclass · Accepted Answer

$\{-1, 1\}$

Vedclass · Answer

(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(4, 0)$.Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.This line is tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.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 focal chord to $y^2 = 16x$ is tangent to $(x - 6)^2 + y^2 = 2$. Then,the possible values of the slope of this chord are:

Similar Questions

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

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to:

The equation of the common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=2$ is $ax+by+2=0$. If $-\frac{a}{b} > 0$,then $3a^2+2b+1=$

The equation of the circle symmetric to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$ about the line $x - y = 3$ is

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. If a straight line $L$ which is perpendicular to $PT$ is a tangent to the circle $(x-3)^2+y^2=1$,then a possible equation of $L$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module