iCON Education - Parabola BITSAT Test

iCON Education iconic preparation for JEE and NEET

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Candidate Name

Student

Exam Name

BITSAT

Test Name

Parabola

Subject

Mathematics

← Back to Parabola

Remaining Time

00:15:00

Question 1: Palette 1/6

BITSAT 2023 MCQ

If $y=m_1 x+c_1$ and $y=m_2 x+c_2, m_1 \neq m_2$ are two common tangents of circle $x^2+y^2=2$ and parabola $y^2=x$, then the value of $8\left|m_1 m_2\right|$ is equal to

$7+6 \sqrt{2}$

$3+4 \sqrt{2}$

$-5+6 \sqrt{2}$

$-4+3 \sqrt{2}$

Question 2: Palette 2/6

BITSAT 2022 MCQ

If the straight line $y = mx + c$ touches the parabola ${y^2} - 4ax + 4{a^3} = 0$, then c is

$am + {a \over m}$

$am - {a \over m}$

${a \over m} + {a^2}m$

${a \over m} - {a^2}m$

Question 3: Palette 3/6

BITSAT 2022 MCQ

A normal is drawn at the point P to the parabola ${y^2} = 8x$, which is inclined at 60$^\circ$ with the straight line $y = 8$. Then the point P lies on the straight line

$2x + y - 12 - 4\sqrt 3 = 0$

$2x - y - 12 + 4\sqrt 3 = 0$

$2x - y - 12 - 4\sqrt 3 = 0$

None of these

Question 4: Palette 4/6

BITSAT 2022 MCQ

For each parabola y = x² + px + q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

(1, 0)

(0, 1)

(1, 1)

(p, q)

Question 5: Palette 5/6

BITSAT 2021 MCQ

The origin is shifted to (1, 2). The equation y² $-$ 8x $-$ 4y + 12 = 0 changes to y² = 4ax, then a is equal to

$-$2

$-$1

Question 6: Palette 6/6

BITSAT 2020 MCQ

The distance of point of intersection of the tangents to the parabola x = 4y $-$ y² drawn at the points where it is meet by Y-axis, from its focus is

${{11} \over 4}$

${{17} \over 4}$

${{13} \over 4}$

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