A parabola S=0 has its vertex at (-9,3) and it touches the x-axis at the origin then equation of axis of symmetry of the aforesaid parabola can be.

Question

A parabola S=0 has its vertex at (-9,3) and it touches the x-axis at the origin then equation of axis of symmetry of the aforesaid parabola can be. (A) x-y+12=0 (B) x-2 y+15=0 (C) 2 x-y+21=0 (D) x+y+6=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/066f2a2e00dd103953c817b7ee837f17-.jpeg" /> V C=3 units, C O=9 units As per property: A V=V N and (A N/A V)=2=(M N/V C) ⇒ O M=6 tan θ, A C=C M=3 cot θ ⇒ 9=3(2 tan θ+(1/ tan θ)) ⇒ m=1 text or m=(1/2) ⇒ Eq n of Axis can be (y-3)=1(x+9) or(y-3)=(1/2)(x+9)

Q. A parabola $S = 0$ has its vertex at $(- 9, 3)$ and it touches the $x$ -axis at the origin then equation of axis of symmetry of the aforesaid parabola can be.

$x - y + 12 = 0$

$x - 2 y + 15 = 0$

$2 x - y + 21 = 0$

$x + y + 6 = 0$

Q. A parabola S=0 has its vertex at (−9,3) and it touches the x-axis at the origin then equation of axis of symmetry of the aforesaid parabola can be.

x−y+12=0

x−2y+15=0

2x−y+21=0

x+y+6=0

VC=3 units, CO=9 units As per property: AV=VN and AVAN​=2=VCMN​ ⇒OM=6tanθ,AC=CM=3cotθ⇒9=3(2tanθ+tanθ1​) ⇒m=1 or m=21​ ⇒Eqn of Axis can be (y−3)=1(x+9) or(y−3)=21​(x+9)

Q. A parabola $S = 0$ has its vertex at $(- 9, 3)$ and it touches the $x$ -axis at the origin then equation of axis of symmetry of the aforesaid parabola can be.

$x - y + 12 = 0$

$x - 2 y + 15 = 0$

$2 x - y + 21 = 0$

$x + y + 6 = 0$