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Application of Derivatives question

2004 · Shift 0 · Q78

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Application of Derivatives
2004 · Shift 0 · Q78

Application of Derivatives question

2004 · Shift 0 · Q78

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1

A point on the parabola

{y^2} = 18x

at which the ordinate increases at twice the rate of the abscissa is

A
$\left( {{9 \over 8},{9 \over 2}} \right)$
B
$(2, -4)$
C
$\left( {{-9 \over 8},{9 \over 2}} \right)$
D
$(2, 4)$

View written solutionFree

Correct answer: A

We are given the parabola $y^2=18x$

We need the point where the ordinate increases at twice the rate of the abscissa.

That means $\frac{dy}{dt}=2\frac{dx}{dt}$ so, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=2$

Differentiate the parabola implicitly with respect to $x$ : $y^2=18x$ $2y\frac{dy}{dx}=18$ $\frac{dy}{dx}=\frac{9}{y}$
Use the condition $\frac{dy}{dx}=2$ : $\frac{9}{y}=2$ $y=\frac{9}{2}$
Substitute this value of $y$ into the parabola equation: $y^2=18x$ $\left(\frac{9}{2}\right)^2=18x$ $\frac{81}{4}=18x$ $x=\frac{81}{72}=\frac{9}{8}$
Therefore, the required point is $\left(\frac{9}{8},\frac{9}{2}\right)$
Check options:

A: $\left(\frac{9}{8},\frac{9}{2}\right)$ ✔
B: $(2,-4)$ gives $\frac{dy}{dx}=\frac{9}{-4}=-\frac{9}{4}\ne 2$
C: $\left(-\frac{9}{8},\frac{9}{2}\right)$ does not satisfy $y^2=18x$
D: $(2,4)$ gives $\frac{dy}{dx}=\frac{9}{4}\ne 2$

Hence the correct option is A.

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