Equation of Parabola MCQ Quiz in मल्याळम - Objective Question with Answer for Equation of Parabola - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Equation of parabola having vertex at origin and along X-axis: y² = 4ax

Calculation:

It is given that the vertex of the parabola is at the origin and its axis lies along the X-axis. So,its equation is \(\rm y^2=4ax\)  OR \(\rm y^2=-4ax\)

Since it passes through the point P(3, 4), so it lies in the first quadrant.

∴ Its equation is \(\rm y^2=4ax\)

Now,  P(3, 4) lies on it, so

\(\rm 4^2=4a(3)=12a\)

\(\rm \Rightarrow a = \frac43\)

∴ The required equation is \(\rm y^2=4(\frac43)x\)

\(\Rightarrow \rm y^2=\frac{16}{3}x\)

Hence, option (4) is correct. 

Solution:

According to the data given above, the equation of the parabola will be y² = 16x

Given that, inclination of the tangent is θ = 60^o

⇒ m = tan60^o = \(\sqrt{3}\)

∴ Equation of the tangent is a \(y = mx+\frac{a}{m}\)

Here, a = 4 and m = \(\sqrt{3}\)

⇒ \(y=\sqrt{3}x+\frac{4}{\sqrt{3}}\)

⇒ \(\sqrt{3}y = 3x+4\)

⇒ \(3x-\sqrt{3}y+4=0\)

∴ The correct option is (1)

Concept:

By definition, the distance between a point on the parabola and its focus is equal to the distance between the point and the directrix (i.e. eccentricity = 1).

The distance between a point (x1, y1) and the line ax + by + c = 0 is given by: d = \(\rm \left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|\)

Calculation:

Let P(x, y) be a point on the parabola, F(-1, -2) be the focus and D = x - 2y + 3 = 0 be the given directrix.

PF = PD

⇒ PF² = PD²

⇒ (x + 1)² + (y + 2)² = \(\rm \left(\frac{x-2y+3}{\sqrt{1+4}}\right)^2\)

⇒ 5[(x + 1)2 + (y + 2)2] = (x - 2y + 3)²

⇒ 5(x2 + 2x + 1 + y2 + 4y + 4) = x² + 4y² + 9 - 4xy + 6x - 12y

⇒ 4x² + y² + 4xy + 4x + 32y + 16 = 0

CONCEPT:

The following are the properties of a parabola of the form: x2 = - 4ay where a > 0

• Focus is given by (0, - a)
• Vertex is given by (0, 0)
• Equation of directrix is given by: y = a
• Equation of axis is given by: x = 0
• Length of latus rectum is given by: 4a
• Equation of latus rectum is given by: y = - a

CALCULATION:

Here, we have to find the equation of the parabola with vertex at origin, passing through the point (3, - 4) and symmetric about the y-axis.

It is given that the vertex of the parabola is at (0, 0) and it is symmetric about the y-axis.

So, the equation of parabola can be either x² = 4ay or x² = - 4ay

But since the parabola passes through the point P (3, - 4) and the point P lies in the 4th quadrant

i.e the parabola is facing downwards

So, the equation of the parabola is of the form x2 = - 4ay

∵ The point P (3, - 4) passes through the parabola so, x = 3 and y = - 4 will satisfy the equation x2 = - 4ay

⇒ 3² = - 4 ⋅ a ⋅ (- 4)

⇒ a = 9/16

So, the required equation of parabola is: x² = - 4 ⋅ (9/16)y 

⇒ 4x² = - 9y

Hence, option A is the correct answer.

CONCEPT:

The following are the properties of a parabola of the form: y2 = 4ax where a > 0

• Focus is given by (a, 0)
• Vertex is given by (0, 0)
• Equation of directrix is given by: x = - a
• Equation of axis is given by: y = 0
• Length of latus rectum is given by: 4a
• Equation of latus rectum is given by: x = a

CALCULATION:

Here, we have to find the equation of the parabola whose focus is at F(3, 0) and directrix x = - 3.

As we know,

Parabola of the form y2 = 4ax has focus at (a, 0) and the equation of directrix is given by x = - a

So, by comparing the focus F(3, 0) and directrix x = - a with (a, 0) and x = - a respectively

we get

⇒ a = 3

So, the equation of the required parabola is y2 = 4× 3x = 12x

Hence, option B is the correct answer.

Given: 

\(\frac{d y}{d x}=\frac{y}{2(x-2)}, x>0, y>0 .\)

Concept: 

The general equation of a parabola where (h, k) denotes the vertex is

y = a (x - h)2 + k, Vertex (h, k), Focus (h, k + (1/4a))

x = a (y - k)2 + h, Vertex (h, k), Focus (h + (1/4a), k)

Calculation:

\(\frac{d y}{d x}=\frac{2y}{(x-2)}\)

Integrating both sides,

\(\int\frac{d y}{y}=\int\frac{2dx}{(x-2)}\)

\({ln\:y}={2ln\:(x-2)}+{ln\:c}\)

\({ln\:y}={ln\:(x-2)^2}+{ln\:c}\)

\({ln\:y}={ln\:c(x-2)^2}\)

\(y = c(x - 2)^2\)

Parabola passes through (3, 2)

So, x = 3, y = 2

2 = c (3 - 2)2

c = 2

The equation of parabola is \(y = 2(x - 2)^2\)

Now comparing the given equation of a parabola with the general form y = a (x - h)2 + k, 

Here, (h, k) is (2, 0), and a = 2

Vertex = (2, 0)​

Focus = (h, k + (1/4a)) = (2, 1/8)

∴ The correct answer is option (3).

Concept:

Equation of normal to a curve at point (h, k) on the curve is

 \(y-k=-\cfrac{1}{\frac{dy}{dx}\big|_{(h,k)}}(x-h)\)

Equation of tangent to a curve at point (h, k) on the curve is

 \(y-k=\cfrac{dy}{dx}\Big|_{(h,k)}(x-h)\) 

Calculation:

Given equation of the parabola y² = 12x  ...(i)

Differentiate both sides with respect to x, 

\(⇒ 2y\cfrac{dy}{dx}=12\)

\(⇒ \cfrac{dy}{dx}=\cfrac{6}{y}\)   ...(ii) 

Equation of normal to a curve at point (h, k) on the curve is

 \(y-k=-\cfrac{1}{\frac{dy}{dx}\big|_{(h,k)}}(x-h)\)

\(\therefore\)Equation of normal to the parabola at point A( 3, - 6) on the curve is

 \(y-(-6)=-\cfrac{1}{-1}(x-3)\)

⇒ y + 6 = x - 3

⇒ x = y + 9

Substituting x in eq (i)

⇒ y² = 12 (y + 9)

⇒ y² - 12y - 108 = 0

⇒ y² - 18 y + 6 y - 108 = 0

⇒ y( y - 18) + 6 (y - 18) = 0

⇒ (y + 6) ( y - 18) = 0

⇒ y = 18, - 6

We take y = 18, substitute y in eq(i)

18² = 12x 

⇒ x = 27

Hence point P where the normal cuts parabola again is (27, 18).

Equation of tangent to a curve at point (h, k) is

 \(y-k=\cfrac{dy}{dx}\big|_{(h,k)}(x-h)\)

We know from eq(ii) that \(\cfrac{dy}{dx}=\cfrac{6}{y}\)

\(\therefore \cfrac{dy}{dx}|_{P(27,18)}=\cfrac{6}{18}=\cfrac{1}{3}\)

Equation of tangent to parabola at point P(27, 18) is 

\(y-18=\cfrac{dy}{dx}|_{(27,18)}(x-27)\)

\(y-18=\frac{1}{3}(x-27)\)

3y - 54 = x - 27

x - 3y + 27 = 0

The correct answer is option (1).

Concept:

The following are the properties of a parabola of the form: y² = 4ax where a > 0

• Focus is given by: (a, 0)
• Vertex is given by: (0, 0)
• Equation of directrix is given by: x = -a
• Equation of axis is given by: y = 0
• Length of latus rectum is given by: 4a

Solution:

According to the data given above, the equation of the parabola will be y2 = 16x

Here, 4a = 16

⇒ a = 4

So, The equation of directrix of the parabola is x = -a

⇒ x = -4

∴ The correct option is (4)

Concept:

• Focus is the mid point of the latus rectum L and L'
• Length of the latus rectum = 4a

Solution:

Given: The end points of the latus rectum of L(4, 8) and L'(4, -8)

Focus is the mid point of the latus rectum L(4, 8) and L'(4, -8)

S = \(\left ( \frac{4+4}{2},\frac{8-8}{2} \right )\) = (4, 0)

Which is the point on the x-axis.

∴ The axis of the parabola is x-axis.

Now, Length of the latus rectum = 8 + 8 = 16 ⇒ 4a = 16

⇒ a = 4

The distance of the focus from the vertex = 4

∴ Vertex is (0, 0)

By the data given, the parabola is open right.

∴ The equation of the parabola with vertex (0, 0) axis along x-axis and open right is y² = 4ax

⇒ y² = 16x

Which is the equation of the required parabola.

∴ The correct option is (2)

CONCEPT:

The following are the properties of a parabola of the form: x2 = - 4ay where a > 0

• Focus is given by (0, - a)
• Vertex is given by (0, 0)
• Equation of directrix is given by: y = a
• Equation of axis is given by: x = 0
• Length of latus rectum is given by: 4a
• Equation of latus rectum is given by: y = - a

CALCULATION:

Here, we have to find the equation of the parabola whose focus is at F(0, - 3) and directrix y = 3.

As we know that, parabola of the form x2 = - 4ay has focus at (0, - a) and equation of directrix is given by y = a.

So, by comparing the focus F(0, - 3) and directrix y = 3 with (0, - a) and y = a respectively we get

⇒ a = 3

So, the equation of the required parabola is x2 = - 4 ⋅ 3 ⋅ y = - 12y

Hence, option C is the correct answer.