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.

Concept:

Focal Chord of Parabola and Circle Properties:

• The focal chord of the parabola \( y^2 = 4ax \)passes through points on the parabola making a certain angle with the x-axis.
• The circle with one diameter as the focal chord touches the y-axis at a certain point.

• The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.
• The equation of the circle with diameter PS and its intersection with the y-axis is used to find \(\alpha\)

Calculation:

Given,

The parabola: \( y^2 = 4x \)

Focal chord PQ makes an angle of \(60^\circ\) with positive x-axis.

Using slope\( m = \tan 60^\circ = \sqrt{3}\)

\( \tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies t = \sqrt{3} \)

Coordinates of P: \( (3, 2\sqrt{3}) \)

Circle with diameter PS and focus S:

\( (x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0 \)

At x = 0,

\( (-1)(-3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0 \implies (y - \sqrt{3})^2 = 0 \implies y = \sqrt{3} = \alpha \)

Calculate \( 5\alpha^2:\)

\( 5 \alpha^2 = 5 (\sqrt{3})^2 = 5 \times 3 = 15 \)

∴ The correct answer is Option 1.

Calculation

For the parabola \(y^2 = 8x \), the equation of the tangent in slope form is:

\(y = mx + \frac{2}{m}, \)

where m is the slope of the tangent.

The tangent must also touch the circle \(x^2 + y^2 = 2 \)

The perpendicular distance of (0, 0) to the tangent \( y = mx + \frac{2}{m} \) should be equal to the radius of the circle √2.

The perpendicular distance d from the origin (0, 0) to the line \(y = mx + \frac{2}{m} \) is given by:

\(d = \frac{\left|0 - 0 + \frac{2}{m}\right|}{√{1 + m^2}} = \frac{\frac{2}{m}}{√{1 + m^2}}\)

⇒ \(\frac{\frac{2}{m}}{√{1 + m^2}} = √{2} \)

⇒ \(\frac{4}{m^2(1 + m^2)} = 2 \)

⇒ \(4 = 2m^2(1 + m^2) \)

⇒ \(2 = m^2(1 + m^2) \)

⇒ \(2 = m^2 + m^4 \)

⇒ \(m^4 + m^2 - 2 = 0. \)

Let \( z = m^2 \)
⇒ \(z^2 + z - 2 = 0 \)

⇒ (z + 2)(z - 1) = 0.

Thus  z = 1  (since z = -2 is not valid for \(m^2 \))

⇒ \(m = \pm 1 \)

Equation of the common tangents

For m = 1, the tangent equation is:

⇒ y = x + 2.

For m = -1, the tangent equation is:

⇒ y = -x + 2.

These are the equations of the two direct common tangents.

Points of tangency on the circle

For y = x + 2:

Substitute y = x + 2 into \(x^2 + y^2 = 2 \)

\(x^2 + (x + 2)^2 = 2 \)

⇒ \(x^2 + x^2 + 4x + 4 = 2 \)

⇒ \(2x^2 + 4x + 2 = 0 \)

⇒ \(x^2 + 2x + 1 = 0 \)

⇒ \((x + 1)^2 = 0 \implies x = -1 \)

Substituting x = -1 into y = x + 2

⇒ y = -1 + 2 = 1

So, the point of tangency is (-1, 1).

For y = -x + 2

Substitute y = -x + 2 into \( x^2 + y^2 = 2 \)

⇒ \(x^2 + (-x + 2)^2 = 2 \)

⇒ \(x^2 + x^2 - 4x + 4 = 2 \)

⇒ \(2x^2 - 4x + 2 = 0 \)

⇒ \(x^2 - 2x + 1 = 0 \)

⇒ \((x - 1)^2 = 0 \implies x = -1 \)

Substituting x = -1 into y = -x + 2

⇒ y = -(-1) + 2 = -1

So, the point of tangency is (-1, -1).

Thus, the points of tangency on the circle are P(-1, 1) and Q(-1, -1)

Points of tangency on the parabola

Substitute the tangent equations into the parabola y^2 = 8x to find the points of tangency:

For y = x + 2:

Substitute y = x + 2 into \(y^2 = 8x \)
   
⇒ \((x + 2)^2 = 8x \)

⇒ \(x^2 + 4x + 4 = 8x \)

⇒ \(x^2 - 4x + 4 = 0 \)

⇒ \((x - 2)^2 = 0 \implies x = 2 \)

Substituting x = 2 into y = x + 2

⇒ y = 2 + 2 = 4

So, the point of tangency is R(2, 4).

For y = -x + 2

Substitute y = -x + 2 into \( y^2 = 8x \)

⇒ \((-x + 2)^2 = 8x \)

⇒ \(x^2 - 4x + 4 = 8x \)

⇒ \(x^2 - 12x + 4 = 0 \)

⇒ \(x = 2, \quad y = -2 \)

So, the point of tangency is S(2, -4)

The vertices of the quadrilateral are

P(-1, 1), Q(-1, -1), R(2, 4), S(2, -4)

PR = \(√{(4-1)^2 +(2+1)^2} \) = 3√2 

Hence Option(2) is the correct answer.

Calculation

Let the equation chord be

\(y=mx+c\)

y² = 6x

⇒ \(4a=6\implies a=1.5\)

Now as mentioned in the question, this line will pass from focus\((1.5,0)\).

Substitute this in equation we get \(1.5m+c=0\) ...(1)

Now Distance of this line from vertex is given to be \(\dfrac { \sqrt { 5 } }{ 2 } \) so using the formula of perpendicular distance of a line from a point, we get \(\dfrac { c }{ \sqrt { { m }^{ 2 }+1 } } =\dfrac { \sqrt { 5 } }{ 2 }\)

Taking square

\(\Rightarrow 4{ c }^{ 2 }=5{ m }^{ 2 }+5\) ...(2)

Solving both the equation we get \(m=\dfrac { \sqrt { 5 } }{ 2 }\)

Hence option 1 is correct

Calculation

For the parabola \(y^2 = 8x \), the equation of the tangent in slope form is:

\(y = mx + \frac{2}{m}, \)

where m is the slope of the tangent.

The tangent must also touch the circle \(x^2 + y^2 = 2 \)

The perpendicular distance of (0, 0) to the tangent \( y = mx + \frac{2}{m} \) should be equal to the radius of the circle √2.

The perpendicular distance d from the origin (0, 0) to the line \(y = mx + \frac{2}{m} \) is given by:

\(d = \frac{\left|0 - 0 + \frac{2}{m}\right|}{√{1 + m^2}} = \frac{\frac{2}{m}}{√{1 + m^2}}\)

⇒ \(\frac{\frac{2}{m}}{√{1 + m^2}} = √{2} \)

⇒ \(\frac{4}{m^2(1 + m^2)} = 2 \)

⇒ \(4 = 2m^2(1 + m^2) \)

⇒ \(2 = m^2(1 + m^2) \)

⇒ \(2 = m^2 + m^4 \)

⇒ \(m^4 + m^2 - 2 = 0. \)

Let \( z = m^2 \)
⇒ \(z^2 + z - 2 = 0 \)

⇒ (z + 2)(z - 1) = 0.

Thus  z = 1  (since z = -2 is not valid for \(m^2 \))

⇒ \(m = \pm 1 \)

Equation of the common tangents

For m = 1, the tangent equation is:

⇒ y = x + 2.

For m = -1, the tangent equation is:

⇒ y = -x + 2.

These are the equations of the two direct common tangents.

Points of tangency on the circle

For y = x + 2:

Substitute y = x + 2 into \(x^2 + y^2 = 2 \)

\(x^2 + (x + 2)^2 = 2 \)

⇒ \(x^2 + x^2 + 4x + 4 = 2 \)

⇒ \(2x^2 + 4x + 2 = 0 \)

⇒ \(x^2 + 2x + 1 = 0 \)

⇒ \((x + 1)^2 = 0 \implies x = -1 \)

Substituting x = -1 into y = x + 2

⇒ y = -1 + 2 = 1

So, the point of tangency is (-1, 1).

For y = -x + 2

Substitute y = -x + 2 into \( x^2 + y^2 = 2 \)

⇒ \(x^2 + (-x + 2)^2 = 2 \)

⇒ \(x^2 + x^2 - 4x + 4 = 2 \)

⇒ \(2x^2 - 4x + 2 = 0 \)

⇒ \(x^2 - 2x + 1 = 0 \)

⇒ \((x - 1)^2 = 0 \implies x = -1 \)

Substituting x = -1 into y = -x + 2

⇒ y = -(-1) + 2 = -1

So, the point of tangency is (-1, -1).

Thus, the points of tangency on the circle are P(-1, 1) and Q(-1, -1)

Points of tangency on the parabola

Substitute the tangent equations into the parabola y^2 = 8x to find the points of tangency:

For y = x + 2:

Substitute y = x + 2 into \(y^2 = 8x \)
   
⇒ \((x + 2)^2 = 8x \)

⇒ \(x^2 + 4x + 4 = 8x \)

⇒ \(x^2 - 4x + 4 = 0 \)

⇒ \((x - 2)^2 = 0 \implies x = 2 \)

Substituting x = 2 into y = x + 2

⇒ y = 2 + 2 = 4

So, the point of tangency is R(2, 4).

For y = -x + 2

Substitute y = -x + 2 into \( y^2 = 8x \)

⇒ \((-x + 2)^2 = 8x \)

⇒ \(x^2 - 4x + 4 = 8x \)

⇒ \(x^2 - 12x + 4 = 0 \)

⇒ \(x = 2, \quad y = -2 \)

So, the point of tangency is S(2, -4)

The vertices of the quadrilateral are

P(-1, 1), Q(-1, -1), R(2, 4), S(2, -4)

PR = \(√{(4-1)^2 +(2+1)^2} \) = 3√2 

Hence Option(2) is the correct answer.

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