Q98 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

If $a x^2+b y^2=15$ is the equation of the ellipse for which distance between its foci is 2 and distance between its directrices is 5 , then $a+b=$

  • A
    10
  • B
    8
  • Correct answer
    C
    16
  • D
    12

Solution

Given equation is $a x^2+b y^2=15$ $$ \frac{x^2}{\left(\sqrt{\frac{15}{a}}\right)^2}+\frac{y^2}{\left(\sqrt{\frac{15}{b}}\right)^2}=1 $$ Now, $a^{\prime}=\sqrt{\frac{15}{a}}, b^{\prime}=\sqrt{\frac{15}{b}}$ According to question, $$ \begin{aligned} & 2 d e=2 \Rightarrow a^{\prime} e=1 \\ & \frac{2 a^{\prime}}{e}=5 \Rightarrow \frac{a^{\prime}}{e}=\frac{5}{2} \end{aligned} $$ Multiply (ii) and (iii), $$ a^{\prime} e \times \frac{a^{\prime}}{e}=1 \times \frac{5}{2} $$ $$ a^{\prime 2}=\frac{5}{2} a^{\prime}= \pm \frac{\sqrt{5}}{\sqrt{2}} $$ from (i) $$ a^{\prime} e=1 $$ $$ \begin{aligned} & e= \pm \frac{\sqrt{2}}{\sqrt{5}} \\ & b^{\prime 2}=a^{\prime 2}=\left(1-e^2\right) \\ & b^{\prime 2}=\frac{5}{2}\left(1-\frac{2}{5}\right)=\frac{5}{2} \times \frac{3}{5}=\frac{3}{2} \\ & b^{\prime}= \pm \frac{\sqrt{3}}{\sqrt{2}} \end{aligned} $$ Now, compare $a^{\prime} \& b^{\prime}$ values, $$ \frac{\sqrt{15}}{\sqrt{a}}= \pm \frac{\sqrt{5}}{\sqrt{2}} $$ square both sides, $$ \frac{15}{a}=\frac{5}{2} \Rightarrow a=6 \text {. } $$ Similarly, $\frac{\sqrt{15}}{\sqrt{b}}= \pm \frac{\sqrt{3}}{\sqrt{2}}$ $$ \begin{aligned} & \frac{15}{b}=\frac{3}{2} \Rightarrow b=10 \\ & \text { So, } a+b=10+6=16 \end{aligned} $$

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

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