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\[

A.\left( {\dfrac{9}{8},\dfrac{9}{2}} \right) \\

B.\left( {2, - 4} \right) \\

C.\left( {\dfrac{{ - 9}}{8},\dfrac{9}{2}} \right) \\

D.\left( {2,4} \right) \\

\]

Answer

Verified

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Hint: Differentiate the given curve equation and equate with the curve equation to find the points.

Given that:

Curve equation ${y^2} = 18x$

Ordinate increases twice the abscissa

So, $\dfrac{{dy}}{{dx}} = 2$ -- (1)

Differentiating the given parabola equation we get

$

2ydy = 18dx \\

\dfrac{{dy}}{{dx}} = \dfrac{{18}}{{2y}} \\

$ --- (2)

From equation 1 and 2, we have

$

\dfrac{{18}}{{2y}} = 2 \\

y = \dfrac{9}{2} \\

$

Substituting the value of $y$ obtained in the given curve equation:

$

\Rightarrow {y^2} = 18x \\

\Rightarrow \dfrac{{81}}{4} = 18x \\

\Rightarrow x = \dfrac{9}{8} \\

$

Hence, the point is $\left( {\dfrac{9}{8},\dfrac{9}{2}} \right)$

Correct answer is option A.

Note:The following curve given in the question represents a parabola about x-axis. The parabola is the locus of points in that plane that are equidistant from both the directrix and the focus.

Given that:

Curve equation ${y^2} = 18x$

Ordinate increases twice the abscissa

So, $\dfrac{{dy}}{{dx}} = 2$ -- (1)

Differentiating the given parabola equation we get

$

2ydy = 18dx \\

\dfrac{{dy}}{{dx}} = \dfrac{{18}}{{2y}} \\

$ --- (2)

From equation 1 and 2, we have

$

\dfrac{{18}}{{2y}} = 2 \\

y = \dfrac{9}{2} \\

$

Substituting the value of $y$ obtained in the given curve equation:

$

\Rightarrow {y^2} = 18x \\

\Rightarrow \dfrac{{81}}{4} = 18x \\

\Rightarrow x = \dfrac{9}{8} \\

$

Hence, the point is $\left( {\dfrac{9}{8},\dfrac{9}{2}} \right)$

Correct answer is option A.

Note:The following curve given in the question represents a parabola about x-axis. The parabola is the locus of points in that plane that are equidistant from both the directrix and the focus.

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