A common tangent to 9x^2 + 16y^2 = 144,y^2 - | vedclass

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

Question

(C) The given equations are:$1$) Ellipse: $9x^2 + 16y^2 = 144 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$2$) Parabola: $y^2 = x - 4$$3$) Circle:

Vedclass · Accepted Answer

$x = 4$

Vedclass · Answer

(C) The given equations are:$1$) Ellipse: $9x^2 + 16y^2 = 144 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$2$) Parabola: $y^2 = x - 4$$3$) Circle: $x^2 + y^2 - 12x + 32 = 0 \Rightarrow (x-6)^2 + y^2 = 4$For the parabola $y^2 = 4a(x-h)$,the tangent is $y = mx + \frac{a}{m} + h$. Here $4a = 1$,so $a = 1/4$ and $h = 4$. Thus,$y = mx + \frac{1}{4m} + 4$.For the circle $(x-6)^2 + y^2 = 2^2$,the distance from center $(6, 0)$ to the line $mx - y + (\frac{1}{4m} + 4) = 0$ is equal to radius $2$:$\frac{|6m - 0 + \frac{1}{4m} + 4|}{\sqrt{m^2 + 1}} = 2$$|6m + \frac{1}{4m} + 4| = 2\sqrt{m^2 + 1}$.Testing $x = 4$ (vertical line,$m 	o \infty$): For the ellipse,$x=4$ is a tangent at $(4, 0)$.For the parabola $y^2 = x-4$,$x=4$ is a tangent at $(4, 0)$.For the circle $(x-6)^2 + y^2 = 4$,at $x=4$,$(4-6)^2 + y^2 = 4$ $\Rightarrow 4 + y^2 = 4$ $\Rightarrow y = 0$. Thus $x=4$ is a tangent at $(4, 0)$.Therefore,$x=4$ is the common tangent.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

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 $(1, 3)$	$IV$. $x=9$

$A$ common tangent to $9x^2 + 16y^2 = 144$,$y^2 - x + 4 = 0$,and $x^2 + y^2 - 12x + 32 = 0$ is:

Similar Questions

Each of the two orthogonal circles $C_1$ and $C_2$ passes through both the points $(2,0)$ and $(-2,0)$. If $y=mx+c$ is a common tangent to these circles,then

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight 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 can represent $L_1$?

The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0)$ and $(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module