Q95 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

The length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a \gt b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$, then $\sqrt{a^2+6 a b+b^2}=$

  • Correct answer
    A
    7
  • B
    $12 \sqrt{2}$
  • C
    $3 \sqrt{5}$
  • D
    11

Solution

$\begin{aligned} & a c=\sqrt{5}...(i) \\ & e=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow e^2=1-\frac{b^2}{a^2} \\ & \frac{5}{a^2}=1-\frac{b^2}{a^2} \Rightarrow 5=a^2-b^2 \text {.[From (i)] } \\ & \Rightarrow b^2=a^2-5....(ii) \\ & \frac{2 b^2}{a}=\frac{8}{3} \Rightarrow \frac{2\left(a^2-5\right)}{a}=\frac{8}{3} \quad \text {(from using (ii)) } \\ & 3 a^2-4 a-15=0 \\ & \Rightarrow 3 a(a-3)+5(a-3)=0 \Rightarrow a=3, a=\frac{-5}{3} \\ & \text {Fo: } a=3, b=2 \\ & \Rightarrow \sqrt{a^2+b^2+6 a b}=\sqrt{9+4+6 \times 6}=\sqrt{49}=7 . \end{aligned}$

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

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