Let y=mx+c, m0 be the focal chord of y^{2}=- | vedclass

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

Question

(A) The equation of the parabola is $y^{2}=-64x$. Comparing with $y^{2}=-4ax$,we get $a=16$. The focus is $(-16, 0)$.Since $y=mx+c$ is a focal chord,i

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) The equation of the parabola is $y^{2}=-64x$. Comparing with $y^{2}=-4ax$,we get $a=16$. The focus is $(-16, 0)$.Since $y=mx+c$ is a focal chord,it passes through $(-16, 0)$,so $0 = m(-16) + c$,which gives $c=16m$.The line $y=mx+c$ is tangent to the circle $(x+10)^{2}+y^{2}=4$. The center of the circle is $(-10, 0)$ and the radius is $r=2$.The perpendicular distance from the center $(-10, 0)$ to the line $mx-y+c=0$ is equal to the radius $r=2$.$\frac{|m(-10)-0+c|}{\sqrt{m^{2}+(-1)^{2}}} = 2$$|c-10m| = 2\sqrt{m^{2}+1}$.Substituting $c=16m$,we get $|16m-10m| = 2\sqrt{m^{2}+1}$,so $|6m| = 2\sqrt{m^{2}+1}$.Since $m>0$,$3m = \sqrt{m^{2}+1}$. Squaring both sides,$9m^{2} = m^{2}+1$,so $8m^{2}=1$,which gives $m=\frac{1}{2\sqrt{2}}$.Then $c = 16m = 16 	imes \frac{1}{2\sqrt{2}} = \frac{8}{\sqrt{2}}$.Finally,$4\sqrt{2}(m+c) = 4\sqrt{2}(\frac{1}{2\sqrt{2}} + \frac{8}{\sqrt{2}}) = 4\sqrt{2}(\frac{1+16}{2\sqrt{2}}) = 2(17) = 34$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

Let $y=mx+c, m>0$ be the focal chord of $y^{2}=-64x$,which is tangent to $(x+10)^{2}+y^{2}=4$. Then,the value of $4\sqrt{2}(m+c)$ is equal to $.....$

Similar Questions

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is ...........

Let $r$ be the radius of the circle,which touches the $x$-axis at point $(a, 0)$,where $a < 0$,and the parabola $y^2 = 9x$ at the point $(4, 6)$. Then $r$ is equal to . . . . . . .

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. Find the equation of the circle with $AB$ as its diameter.

Let $L_1$ be a line passing through the origin and $L_2$ be the line $x + y = 1$. If the intercepts made by the circle $x^{2} + y^{2} - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations represents $L_1$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module