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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)=∫01x2+t22−t dty=f(x)=\displaystyle\int_{0}^{1}\dfrac{x^{2}+t^{2}}{2-t}\,dt. It represents
Aa parabola with latus rectum ln⁡2\ln 2
Ban ellipse with latus rectum ln⁡4\ln 4
Ca parabola with latus rectum 1ln⁡2\dfrac{1}{\ln 2}correct
Dan ellipse with latus rectum 1ln⁡2\dfrac{1}{\ln 2}
Solution
Step 1: The integration is over tt, so xx is a parameter that can be pulled outside. Separate the integrand:
y=∫01x22−t dt+∫01t22−t dt=x2∫01dt2−t+∫01t22−t dt.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 x2x^{2}:
∫01dt2−t=[−ln⁡(2−t)]01=−ln⁡1+ln⁡2=ln⁡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 cc. Then
y=(ln⁡2) x2+c,y=(\ln 2)\,x^{2}+c,
 which is a parabola opening upward in the xyxy-plane. Step 4: Rewrite in standard form x2=1ln⁡2 (y−c)x^{2}=\dfrac{1}{\ln 2}\,(y-c). Comparing with X2=4a YX^{2}=4a\,Y, the latus rectum is 4a=1ln⁡24a=\dfrac{1}{\ln 2}. Correct answer: (3)
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