Vertex of the parabola $9{x^2} - 6x + 36y + 9 = 0$ is

$\left( { - 1/3, - 1/2} \right)$

The focus of the parabola $y=2 x^2+x$ is

$\left(\frac{1}{2}, \frac{1}{4}\right)$

$\left(-\frac{1}{4}, \frac{1}{8}\right)$

The vertex of the parabola ${x^2} + 2y = 8x - 7$ is

$\left( {\frac{9}{2},0} \right)$

$\left( {2,\frac{9}{2}} \right)$

$\left( {4,\frac{7}{2}} \right)$

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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-390 of 586 questions

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361 Easy

361 of 586

Application +4 | -1

The point on the curve

{y^2} = ax

the tangent at which makes an angle of

{45^0}

with

x

-axis will be given by

View answer and explanation

Correct option: C

\left( {a/4,a/2} \right)

Option C (

\left( {a/4,a/2} \right)

) is correct because it represents the point on the parabola where the tangent makes a 45-degree angle with the x-axis, calculated using the tangent's slope.

362 Medium

362 of 586

Application +4 | -1

Common tangent is drawn to

y^2=16 x

and

x^2+y^2=8

. Find square of distance between point of contact of common tangent on both the curves.

View answer and explanation

Correct option: B

Option B (72) is correct because it is the square of the distance between the points of contact of the common tangent on the parabola and the circle, calculated using the properties of tangents and circles.

363 Easy

363 of 586

Application +4 | -1

The focal distance of a point on the parabola

y^2=16 x

, whose ordinate is twice the abscissa, is

View answer and explanation

Correct option: B

Option B (8) is correct because substituting the ordinate-abscissa relationship into the parabola equation and calculating the focal distance yields 8.

364 Easy

364 of 586

Application +4 | -1

Vertex of the parabola

9{x^2} - 6x + 36y + 9 = 0

View answer and explanation

Correct option: A

\left( {1/3, - 2/9} \right)

Option A ( (1/3, -2/9) ) is correct because it represents the vertex obtained after rewriting the given parabola equation in standard form.

365 Easy

365 of 586

Application +4 | -1

The axis of the parabola

9 y^2-16 x-12 y-57=0

View answer and explanation

Correct option: A

3 y=2

Option A (

3 y=2

) is correct because it represents the axis of symmetry of the given parabola after completing the square.

366 Medium

366 of 586

Analyzing +4 | -1

Two parabolas have the same focus. If their directrices are the

X

-axis and the

Y

-axis, respectively, then the slope of their common chord is

View answer and explanation

Correct option: A

\pm 1

Option C (3) is correct because solving for the intersection points and using the tangent line equation leads to a cubic equation with a = 3 as a valid solution.

367 Easy

367 of 586

Application +4 | -1

The slopes of tangents drawn from a point

(4,10)

to the parabola

y^2=9 x

, are

View answer and explanation

Correct option: B

1 / 4,9 / 4

Option B (

1/4, 9/4

) is correct because these are the slopes of the tangents found by substituting the point (4, 10) into the tangent equation of the parabola.

368 Easy

368 of 586

Understanding +4 | -1

Eccentricity of the parabola

x^2-4 x-4 y+4=0

View answer and explanation

Correct option: B

e=1

Option B (

e=1

) is correct because the eccentricity of a parabola is always equal to 1.

369 Medium

369 of 586

Application +4 | -1

Equation of the common tangent to the circle

x^2+y^2-6 x=0

and the parabola

y^2=4 x

, is

View answer and explanation

Correct option: C

\sqrt{3} y=x+3

Option C (

\sqrt{3} y=x+3

) is correct because it is the tangent to both the circle and the parabola, found by equating the conditions for tangency.

370 Medium

370 of 586

Application +4 | -1

Locus of middle point of the chord of the parabola

y^2=4 x

which passes through a point

(1,2)

View answer and explanation

Correct option: B

(y-1)^2=(2 x-1)

Option B (

(y-1)^2=(2 x-1)

) is correct because the locus of the midpoint of the chord is found using T=S1 and the condition that the chord passes through (1,2).

371 Easy

371 of 586

Analyzing +4 | -1

y = mx + 4

is a tangent to both the parabolas,

{y^2} = 4x

and

\mathrm{ }{x^2} = 2by,

then

b

is equal to

View answer and explanation

Correct option: B

-128

Option B (-128) is correct because by finding the tangent to the first parabola and then using the condition of tangency for the second parabola, we can solve for the value of b.

372 Easy

372 of 586

Application +4 | -1

The value of

c

, for which the line

y = 2x + c

, is tangent to the parabola

{y^2} = 4a\left( {x + a} \right)

, is

View answer and explanation

Correct option: D

\frac{{5a}}{2}

Option D (

\frac{{5a}}{2}

) is correct because it is found by substituting the line equation into the parabola equation and setting the discriminant to zero, then solving for

c

373 Easy

373 of 586

Application +4 | -1

Equation of the directrix of parabola

2{x^2} = 14y

is equal to

View answer and explanation

Correct option: A

y = - \frac{7}{4}

Option A (

y = - \frac{{7}}{{4}}

) is correct because the equation of the directrix of the parabola is

y = -a

, where

a = \frac{7}{4}

374 Easy

374 of 586

Analyzing +4 | -1

The graph represented by the equations

x = {\sin ^2}t,y = 2\cos t,

View answer and explanation

Correct option: B

A parabola

Option B (A parabola) is correct because eliminating the parameter t results in an equation of the form

{y^2} + 4x = 4

, which represents a parabola.

375 Easy

375 of 586

Application +4 | -1

In the parabola

{y^2} = 4ax

, the length of the chord passing through the vertex inclined to the axis at

\frac{\pi }{4}

View answer and explanation

Correct option: A

4a\sqrt 2

Option A (

4a\sqrt 2

) is correct because it represents the length of the chord passing through the vertex of the parabola at an angle of

\frac{\pi}{4}

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376 Medium

376 of 586

Application +4 | -1

The focus of the parabola

y=2 x^2+x

View answer and explanation

Correct option: C

\left(-\frac{1}{4}, 0\right)

Option D (-1/4, 0) is correct because the focus of the parabola is found by rewriting the equation in standard form and applying the formula for the focus.

377 Medium

377 of 586

Remembering +4 | -1

The vertex of the parabola

(y-2)^2=16(x-1)

View answer and explanation

Correct option: D

(1,2)

Option D ((1,2)) is correct because the vertex of the parabola

(y-k)^2=4a(x-h)

is (h, k), and in this case, h=1 and k=2.

378 Easy

378 of 586

Analyzing +4 | -1

Let

l_1

and

l_2

be the two lines which are normal to

y^2=4 x

and tangent to

y^2=-12 y

respectively (where

l_1

and

l_2

are not the

{x}

-axis ). Then, the product of the slopes of

l_1

and

l_2

View answer and explanation

Correct option: B

Option B (2) is correct because solving for the slopes of the normal and tangent lines leads to the product of the slopes being 2.

379 Easy

379 of 586

Application +4 | -1

If the focus and vertex of a parabola are the points

\left( {0,2} \right)

and

\left( {0,4} \right)

respectively, then its equation is

View answer and explanation

Correct option: C

{x^2} + 8y = 32

Option C (

{x^2} + 8y = 32

) is correct because it represents a parabola with the given focus and vertex, derived by using the definition of a parabola and its directrix.

380 Easy

380 of 586

Application +4 | -1

The vertex of the parabola

{x^2} + 2y = 8x - 7

View answer and explanation

Correct option: B

\left( {4,\frac{9}{2}} \right)

Option B (

(4,\frac{9}{2})

) is correct because it represents the vertex of the parabola after rewriting the given equation in the standard form.

381 Medium

381 of 586

Application +4 | -1

If the parabola

x^2= ay

makes an intercept of length

\sqrt{40}

on the line

y-2 x=1

, then

a=

View answer and explanation

Correct option: B

-2

Option B (-2) is correct because solving the quadratic equation derived from the intersection of the parabola and line yields a=-2 as a valid solution.

382 Easy

382 of 586

Application +4 | -1

The parametric coordinates of any point on the parabola whose focus is

(0,1)

and the directrix is

x+2=0

, are

View answer and explanation

Correct option: D

\left(t^2-1,2 t+1\right)

Option D (

t^2-1,2 t+1

) is correct because these parametric coordinates satisfy the equation of the parabola with focus (0,1) and directrix x+2=0.

383 Easy

383 of 586

Application +4 | -1

The line

y=m x+1

is a tangent to the parabola

y^2=4 x

, if

m=

View answer and explanation

Correct option: A

Option A (1) is correct because substituting the line equation into the parabola equation and applying the tangency condition (c = a/m) yields m = 1.

384 Easy

384 of 586

Application +4 | -1

The equation of the chord of

y^2=8 x

which is bisected at

(2,-3)

, is

View answer and explanation

Correct option: A

4 x+3 y+1=0

Option A (

4 x+3 y+1=0

) is correct because the equation of the chord bisected at a given point is found using the formula T = S', where T and S' are calculated based on the given parabola and the point of bisection.

385 Easy

385 of 586

Application +4 | -1

The line

2 x+y+\lambda=0

is a normal to the parabola

y^2=-8 x

, if

\lambda=

View answer and explanation

Correct option: C

Option C (24) is correct because substituting the equation of the normal into the parabola equation and solving for lambda gives 24.

386 Medium

386 of 586

Application +4 | -1

The length of the latusrectum of the parabola

169\left\{ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right\} = {\left( {5x - 12y + 17} \right)^2}

View answer and explanation

Correct option: C

28/13

Option C (

28/13

) is correct because the length of the latusrectum of the parabola is twice the distance of the focus from the directrix.

387 Medium

387 of 586

Application +4 | -1

The area of the triangle formed by tangents and the chord of contact from

(3,4)

y^2=2 x

View answer and explanation

Correct option: A

10 \sqrt{10}

Option A (

10 \sqrt{10}

) is correct because the area of the triangle formed by tangents and the chord of contact is calculated using the formula

\frac{{(S_{11})^{\frac{{3}}{{2}}}}}}{{2a}}

388 Easy

388 of 586

Application +4 | -1

The equation

x^2-2 x y+y^2+3 x+2=0

represents

View answer and explanation

Correct option: A

A parabola

Option A (A parabola) is correct because the discriminant

h^2 - ab = 0

indicates that the equation represents a parabola.

389 Easy

389 of 586

Application +4 | -1

The equation of the tangent to the parabola

y^2=4 a x

at point

\left(a / t^2, 2 a / t\right)

View answer and explanation

Correct option: A

t y=x t^2+a

Option B (

ty=x+at^2

) is correct because it is the equation of the tangent to the parabola

y^2=4ax

at the point

(\frac{a}{t^2}, \frac{2a}{t})

390 Easy

390 of 586

Analyzing +4 | -1

Two perpendicular tangents

P A

and

P B

are drawn to

y^2=4 a x

, minimum length of

A B

is equal to

View answer and explanation

Correct option: C

4 a

Option C (

4a

) is correct because the minimum length of AB is the length of the latus rectum, which is

4a

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