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Curve y=∫(x²+t²)/(2−t)dt: parabola latus rectum | JEE

JEE Maths question with a full step-by-step solution.

Question

Consider the real-valued function

y=f(x)=\displaystyle\int_{0}^{1}\dfrac{x^{2}+t^{2}}{2-t}\,dt

. It represents

Aa parabola with latus rectum

\ln 2

Ban ellipse with latus rectum

\ln 4

Ca parabola with latus rectum

\dfrac{1}{\ln 2}

correct

Dan ellipse with latus rectum

\dfrac{1}{\ln 2}

Solution

Step 1: The integration is over

t

, so

x

is a parameter that can be pulled outside. Separate the integrand:

y=\int_{0}^{1}\dfrac{x^{2}}{2-t}\,dt+\int_{0}^{1}\dfrac{t^{2}}{2-t}\,dt=x^{2}\int_{0}^{1}\dfrac{dt}{2-t}+\int_{0}^{1}\dfrac{t^{2}}{2-t}\,dt.

Step 2: Evaluate the coefficient of

x^{2}

\int_{0}^{1}\dfrac{dt}{2-t}=\big[-\ln(2-t)\big]_{0}^{1}=-\ln 1+\ln 2=\ln 2.

Step 3: The second integral is a finite constant; denote it

c

. Then

y=(\ln 2)\,x^{2}+c,

which is a parabola opening upward in the

xy

-plane. Step 4: Rewrite in standard form

x^{2}=\dfrac{1}{\ln 2}\,(y-c)

. Comparing with

X^{2}=4a\,Y

, the latus rectum is

4a=\dfrac{1}{\ln 2}

. Correct answer: (3)

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