One of the points on the parabola y²=12 x with focal distance 12, is

Correct Answer is : (9,63)(9,6 {3})(9,63)

Correct Answer is : (9,63)(9,6 {3})(9,63)

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KEAM 2010 Math Paper

Question : 1 of 120

Marks: +1, -0

y^{2}=12 x

with focal distance 12, is

(3,6)
$(9,6 \sqrt{3})$
$(7,2 \sqrt{21})$
$(8,4 \sqrt{6})$
$(1, \sqrt{12})$

Solution:

We have

y^{2}=12 x

........(1)

Comparing Eq. (i) with

y^{2}=4 a x,

we get

4a=12

\Rightarrow a=3

∴ Focus

=(3,0)

Let

(x_{1}, y_{1})

be point on parabola Eq. (i), hence

y_{1}^{2}=12 x_{1}

.....(ii)

Now, according to question,

\sqrt{(x_{1}-3)^{2}+(y_{1}-0)^{2}}=12

\Rightarrow \;\; x_{1}^{2}+9-6 x_{1}+y_{1}^{2}=144

....(iii)

Putting

y_{1}^{2}=12 x,

from Eq. (ii) in Eq. (iii), we get

x_{1}^{2}-6 x_{1}+9+12 x_{1}=144

\Rightarrow \;\; x_{1}^{2}+6 x_{1}-135=0

\Rightarrow \;\; (x_{1}+15)(x_{1}-9)=0

\Rightarrow \;\; x_{1}=9,-15

x_{1}=-15

is not possible.

From Eq. (i)

y_{1}^{2}=12 x_{1}

\Rightarrow y_{1}^{2}=12 \times 9=108

\Rightarrow y_{1}=6 \sqrt{3}

∴ Required point

=(9,6 \sqrt{3})

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