Parabola MCQ Quiz - Objective Question with Answer for Parabola - Download Free PDF

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

\(m \;=\; \tan(45^\circ) \;=\; 1.\)

A standard parametric form for y2 = 4x is 

\(\bigl(x(t),\,y(t)\bigr) \;=\;\bigl(t^{2},\,2t\bigr), \)

since  \(y^{2} = 4x \implies (2t)^{2} = 4\,t^{2} \)

The slope of the tangent at \(\bigl(t^{2},\,2t\bigr) \)

Differentiate y2 = 4x 

\(2y\,\frac{dy}{dx} \;=\; 4 \)

\(\;\Longrightarrow\; \frac{dy}{dx} \;=\; \frac{4}{\,2y\,} \;=\; \frac{2}{\,y\,}. \)

At the point \(\bigl(x,y\bigr) = \bigl(t^{2},\,2t\bigr) \) one has \(y = 2t \)

\(\left.\frac{dy}{dx}\right|_{\,(t^{2},\,2t)} \) \(= \frac{2}{\,2t\,} = \frac{1}{t}.\)

We require this slope to equal 1.

\(\frac{1}{t} \;=\; 1 \;\Longrightarrow\; t = 1. \)

Now point of contact

Substitute t = 1 in \(\bigl(x(t),\,y(t)\bigr) = \bigl(t^{2},\,2t\bigr)\). 

\(x(1) = 1^{2} = 1, \)

\(y(1) = 2 \cdot 1 = 2. \)

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

Calculation: 

\(\rm ​​\left(x+\frac{1}{2}\right)^{2}=\left(y+\frac{1}{4}\right) \)

⇒ y = (x² + x)

⇒ tan^-1\(\rm \sqrt{x(x+1)}\) + sin^-1\(\rm \sqrt{x^{2}+x+1}\) = π/2

⇒ 0 ≤ x² + x + 1 ≤ 1

⇒ \(\rm x^{2}+x \leq 0 \quad \quad ...(1)\)

Also \(\rm x^{2}+x \geq 0 \quad \quad ...(2)\)

∴ x² + x = 0 ⇒ x = 0, -1 

S contains 2 element. 

Hence, the correct answer is Option 1. 

Answer (17)

Sol. 

y² = 16x 

⇒ 4a = 16 ⇒ a = 4

Q ≡ \(\left(a t_{2}^{2}, 2 a t_{2}\right)\)

≡ \(\left(4 t_{2}^{2}, 8 t_{2}\right)\)

P ≡ \(\left(4 t_{1}^{2}, 8 t_{1}\right)\)

\(4 t_{1}^{2}\) = 1, 8t₁ = -4 ⇒ t₁ = \(\frac{-1}{2}\)

since P and Q are ends points of focal chord 

t1t2 = 1 ⇒ t2 = 2

⇒ Q = (16,16)

⇒ PF = \(\sqrt{3^{2}+4^{2}}\), FQ = \(\sqrt{12^{2}+16^{2}}\)

⇒ \(\frac{P F}{Q F}=\frac{5}{20}=\frac{1}{4}=\frac{m}{n}\)

⇒ m² + n² = 17 

Calculation: 

The given parabolas are symmetric about the line y = x.

Tangents at A & B must be parallel to y= x line, so the slope of the tangents = 1  

\(\left( \frac{dy}{dx} \right)_{\min A} = 1 = \left( \frac{dy}{dx} \right)_{\min B}\)

For the Point y = x² + 2

^{\(\frac{dy}{dx} = 2x = 1\)}

^{\(x = \tfrac12,\quad y = \bigl(\tfrac12\bigr)^2 + 2 = \tfrac14 + 2 = \tfrac94.\)}

Thus, Point B = ( 1/2, 9/4) ⇒ Point A = (9/4, 1/2)

\(AB = \sqrt{(\frac{1}{2} - \frac{9}{4})^2+ (\frac{9}{4} - \frac{1}{2})^2}\)

AB = \(\sqrt{(\frac{2-9}{4})^2+(\frac{9-2}{4})^2}\)

AB = \(\sqrt\frac{98}{4} = \frac{7\sqrt2 }{4}\) 

Radius = AB/2 = \(\frac{7 \sqrt2}{8}\)

Hence, the correct answer is Option 4.

Concept:

Calculation:

Comparing the given equation (y - 3)2 = 20(x - 1) with the general equation of the parabola (y - k)2 = 4a(x - h), we can say that:

k = 3, a = 5, h = 1.

Vertex is (h, k) = (1, 3).

Concept:

The coordinates of the point where the chord cut the parabola satisfIes  the equation of a parabola.

Calculation:

Given:

The equation of a parabola is y² = x.

The angle made by Chord OA with x-axis is θ

Let the length of the chord OA of the parabola is L 

So, Length of AM = L sinθ 

and Length of OM = L cosθ 

So, The coordinate of A = (L cos θ, L sin θ)

And this point will satisfy the equation of parabola y2 = x.

⇒ (Lsin θ)² = L cos θ

⇒L² sin² θ = L cos θ

⇒ L = cos θ. cosec2 θ

∴ The required length of chord is cos θ. cosec2 θ.

Concept:

Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:

Given: x2 = 16y

⇒ x2 = 4 × 4 × y

Compare with standard equation of parabola x2 = 4ay 

So, a = 4

Therefore, Focus  = (0, a) = (0, 4)

Concept:

Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Given: Equation of hyperbola is x2 - y2 =  1

As we can see that, the given hyperbola is a horizontal hyperbola.

So, by comparing the given equation of hyperbola with \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) we get

⇒ a = 1 and b = 1

As we know that, length of latus rectum of a hyperbola is given by \(\frac{2b^2}{a}\)

So, the length of latus rectum of given hyperbola is 2 units.

Hence, option D is the correct answer.

Concept:

Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:

Given: y2 = -12x

⇒ y2 = 4 × (-3) × x

Compare with standard equation of parabola y² = 4ax

So, a = -3

Therefore, Focus  = (a, 0) = (-3, 0)

Concept:

Latus rectum:

The latus rectum of a conic section is the chord (line segment) that passes through the focus, is perpendicular to the major axis, and has both endpoints on the curve.

• Length of Latus Rectum of Parabola y2 = 4ax is 4a
• End points of the latus rectum of a parabola are L = (a, 2a), and L’ = (a, -2a)

Calculation:

Given equation:

y2 − 8x + 6y + 1 = 0

⇒ y2 + 6y + 9 - 9 - 8x + 1 = 0

⇒ (y + 3)2 - 8x - 8 = 0

⇒ (y + 3)2 = 8x + 8

⇒ (y + 3)2 = 8 (x + 1)

Let new coordinate axes be X and Y,

Here X = x + 1 and Y = y + 3

⇒ Y2 = 4aX

Now comparing with above equation,

∴ 4a = 8 ⇒ a = 2

Focus: (a, 0)

X = a  and Y = 0

⇒ x + 1 = 2 and y + 3 = 0

⇒ x = 1 and y = -3

∴ focus of parabola is (1, -3)

Concept:

The length of the latus rectum of the parabola y2 = 4ax is 4a.

Calculations:

Given, the parabola y2 = 4kx passes through point (-2, 1),

⇒The  point (-2, 1) is satisfying the equation of parabola y2 = 4kx 

⇒ (1)² = 4k (-2)

⇒ k = \(\rm \dfrac {-1}8\)

Now, the length of the latus rectum = 4k

⇒The length of latus rectum = 4(\(\rm \frac {-1}8\))

⇒The length of latus rectum = \(\frac {-1}{2}\)

The length of latus rectum can not be negative.

⇒The length of latus rectum = \(\frac {1}{2}\)

Hence, if parabola y2 = 4kx passes through the point (-2, 1), then the length of the latus rectum is \(\frac {1}{2}\).

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:

Given: Equation of parabola is y2 = - 12x

The given equation can be re-written as: y2 = - 4 ⋅ 3 ⋅ x---------(1)

Now by comparing the equation (1), with y2 = - 4ax we get

⇒ a = 3

As we know that, the length of latus rectum of a parabola is given by: 4a

So, length of latus rectum of the given parabola is: 4 ⋅ 3 = 12 units

Hence, option B is the correct answer.

Concept:

Parabola: The locus of a point which moves such that its distance from a fixed point is equal to its distance from a fixed straight line. (Eccentricity = e =1)

Calculation:

Given: x2 = 20y

⇒ x2 = 4 × 5 × y

Compare with standard equation of parabola x2 = 4ay 

So, 4a = 4 × 5

Therefore, Length of Latus rectum = 4a = 4 × 5 = 20