Q97 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

Let $\mathrm{X}$ - axis be the transverse axis and $\mathrm{Y}$-axis be the conjugate axis of a hyperbola $\mathrm{H}$. Let $\mathrm{x}^2+\mathrm{y}^2=16$ be the director circle of $\mathrm{H}$. If the perpendicular distance from the centre of $\mathrm{H}$ to its latus rectum is $\sqrt{34}$ then $\mathrm{a}+\mathrm{b}=$

  • Correct answer
    A
    8
  • B
    9
  • C
    5
  • D
    7

Solution

We have equation of the director circle of hyperbola $$ \begin{aligned} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } x^2+y^2=a^2-b^2 \\ & \therefore a^2-b^2=16 \end{aligned} $$ We have the perpendicular distance from centre $(0,0)$ of hyperbola to its latus rectum is ae. $$ \begin{aligned} & \therefore \mathrm{ae}=\mathrm{c}=\sqrt{34} \Rightarrow \mathrm{c}^2=34 \\ & \therefore \mathrm{b}^2+\mathrm{a}^2=\mathrm{c}^2=34 \end{aligned} $$ solving equations (i) and (ii) we get $a=5 \& b=3$ $$ \therefore a+b=5+3=8 \text {. } $$

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

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