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GUJCET 2026 Math Paper

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Question : 19 of 40

Marks: +1, -0

If

x = at^2

y = 2at

\frac{d^2y}{dx^2} = \dots

$-\frac{a}{xy}$
$\frac{a}{xy}$
$\frac{ax}{y}$
$-\frac{ax}{y}$

Solution:

Concept:

For parametric curves, use

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

, then differentiate again using the chain rule.

Explanation:

Given

x=at^2

and

y=2at

.

Differentiate with respect to

t

:

\frac{dx}{dt}=2at

,

\frac{dy}{dt}=2a

.

So,

\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}

.

Now,

\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{t}\right)=\frac{d}{dt}\left(\frac{1}{t}\right)\cdot \frac{dt}{dx}

.

Since

\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{2at}

,

\frac{d^2y}{dx^2}=-\frac{1}{t^2}\cdot \frac{1}{2at}=-\frac{1}{2at^3}

.

From

x=at^2

,

t^2=\frac{x}{a}

.

From

y=2at

,

t=\frac{y}{2a}

.

Substitute these:

\frac{d^2y}{dx^2}=-\frac{1}{2a\left(\frac{x}{a}\right)\left(\frac{y}{2a}\right)}=-\frac{a}{xy}

.

Answer:

\frac{d^2y}{dx^2}=-\frac{a}{xy}

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