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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Parabola

BITSAT 2023 · Paper 1 · Mathematics · Parabola

The equation of a common tangent to the parabolas y=x^2 and...

The equation of a common tangent to the parabolas y=x2y=x^2 and y=−(x−2)2y=-(x-2)^2 is

Options

  1. Ay=4(x−2)y=4(x-2)
  2. By=4(x−1)y=4(x-1)
  3. Cy=4(x+1)y=4(x+1)
  4. Dy=4(x+2)y=4(x+2)

Correct answer

B. y=4(x−1)y=4(x-1)

Step-by-step solution

Equation of tangent of parabola y=x2y=x^2 be
tx=y+at2t x=y+a t^2...(i)
y=tx−t24y=t x-\frac{t^2}{4}
Solve with y=−(x−2)2\mathrm{y}=-(\mathrm{x}-2)^2
tx−t24=−(x−2)2t x-\frac{t^2}{4}=-(x-2)^2
x2+x(t−4)−t24+4=0x^2+x(t-4)-\frac{t^2}{4}+4=0
Here, Discriminant =0=0.
(t−4)2−4⋅(4−t24)=0⇒t2−4t=0⇒t=0(t-4)^2-4 \cdot\left(4-\frac{t^2}{4}\right)=0 \Rightarrow t^2-4 t=0 \Rightarrow t=0
or t=4\mathrm{t}=4
Put value of tt in eq. (i), then y=4(x−1)y=4(x-1).
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