Q95 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

If the tangent drawn at the point $P(4,8)$ to the parabola $y^2=16 x$ meets the parabola $y^2=16 x+80$ at $A$ and $B$, then the mid-point of $A B$ is

  • A
    $(9,6)$
  • Correct answer
    B
    $(4,8)$
  • C
    $(4,1)$
  • D
    $(2,3)$

Solution

Equation of tangent to the parabola $y^2=16 x$ at point $P(4,8)$ is $8 y=8(x+4) \Rightarrow y=x+4$ $\ldots(\mathrm{i})$ $\because$ Tangent (i) meets the another parabola $y^2=16 x+80$ at $A$ and $B$, so on solving them we have $(x+4)^2=16 x+80$ $\Rightarrow \quad x^2+8 x+16=16 x+80$ $\Rightarrow(x-4)^2=80 \Rightarrow x=4 \pm 4 \sqrt{5}$ and $y=8 \pm 4 \sqrt{5}$ $\therefore A(4+4 \sqrt{5}, 8+4 \sqrt{5})$ and $B(4-4 \sqrt{5}, 8-4 \sqrt{5})$ $\therefore$ Mid-point of $A B$ is $(4,8)$.

Hence, the correct answer is option: 'B'.

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