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Standard XII

Mathematics

Axis

The vertex of...

Question

The vertex of the parabola ${(y - a)}^{2} = 4 a (x + a)$ is

(- a, a)

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(a, - a)

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(- 2 a, 2 a)

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(- \frac{a}{2}, \frac{a}{2})

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Solution

The correct option is A $(- a, a)$
We know, vertex of parabola $Y^{2} = 4 a X$ is $(X, Y) = (0, 0)$
Hence vertex of parabola, ${(y - a)}^{2} = 4 a (x + a)$
is $(x + a, y - a) = (0, 0)$
$\Rightarrow (x, y) = (- a, a)$

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Similar questions

If normal at p to parabola $y^{2} = 4 a x$ meets parabola again at Q such that PQ subtends a right angle at the vertex of $y^{2} = 4 a x$ then p will be

Q. Assertion :The locus of the center of the circle described on any focal chord of a parabola

y^{2} = 4 a x

as diameter is

y^{2} = 2 a (x - a)

Reason: If

A (a t_{1}^{2}, 2 a t_{1})

and

B (a t_{2}^{2}, 2 a t_{2})

be the extremities of a focal chord for the parabola

y^{2} = 4 a x,

then

t_{1} t_{2} = - 1

Q. Let

f (x) = 4 x^{2} - 4 a x + a^{2} - 2 a + 2

be a quadratic polynomial in

x

a \in R

. If x-coordinate of vertex of parabola

y = f (x)

is less than

0

and

f (x)

has minimum value

3

for

x \in [0, 2]

, then the value of

a

Q. Show that the tangent to

y^{2} = 4 a x

(a, 2 a)

touches the parabola

x^{2} = 4 a y

Q. The equation of tangent drawn from the point

A (2 a, 3 a)

to the parabola

y^{2} = 4 a x

is?

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The vertex of the parabola (y−a)2=4a(x+a) is

The correct option is A (−a,a)We know, vertex of parabola Y2=4aX is (X,Y)=(0,0) Hence vertex of parabola, (y−a)2=4a(x+a)is (x+a,y−a)=(0,0)⇒(x,y)=(−a,a)

The vertex of the parabola ${(y - a)}^{2} = 4 a (x + a)$ is

The correct option is A $(- a, a)$
We know, vertex of parabola $Y^{2} = 4 a X$ is $(X, Y) = (0, 0)$
Hence vertex of parabola, ${(y - a)}^{2} = 4 a (x + a)$
is $(x + a, y - a) = (0, 0)$
$\Rightarrow (x, y) = (- a, a)$