Test Index

Parabola Part 2

Section: Mathematics

Question : 30 of 51

Marks: +1, -0

If the line

3x-2y+12=0

4y

=3x^2

A

B

A B

Solution:

Line

L: 3x-2y+12=0

Parabola

P

4y=3x^2

By putting

y = \frac{3x^2}{4}

in equation of line

We get,

3x-2\left(\frac{3x^2}{4}\right)+12=0

\Rightarrow 6x-3x^2+24=0

\Rightarrow x^2-2x+8=0

\Rightarrow x=4,-2

for

x=4

, we get

y=12

for

x=-2

, we get

y=3

So, points

A

and

B

are

(4,12)

and

(-2,3)

Now, Vertex of parabola is

(0,0)

\Rightarrow \tan\theta = \left|\frac{3-\left(\frac{-3}{2}\right)}{1+3\left(\frac{-3}{2}\right)}\right|

\tan\theta = \frac{9}{7}

\Rightarrow \theta = \tan^{-1}\left(\frac{9}{7}\right)

\Rightarrow

Option (1) is correct

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