The line \(3 x+4 y-5=0\) cuts the curve \(2 x^2+3 y^2=5\) at \(A\) and \(B\). If ' \(O\) ' is the origin,…

The line \(3 x+4 y-5=0\) cuts the curve \(2 x^2+3 y^2=5\) at \(A\) and \(B\). If ' \(O\) ' is the origin, then \(\angle A O B=\)
  1. \(\frac{\pi}{6}\)
  2. \(\frac{\pi}{3}\)
  3. \(\frac{\pi}{2}\)
  4. \(\frac{\pi}{8}\)

Solution

Given equation of line is \(\begin{aligned} 3 x+4 y-5 & =0 \\ \frac{3 x+4 y}{5} & =1 \quad \ldots (i) \end{aligned}\) Equation of curve is \(2 x^2+3 y^2=5\) ...(ii)
Homogenising Eq. (ii) using Eq. (i) \(\begin{aligned} & 2 x^2+3 y^2=5(1)^2 \\ & \Rightarrow \quad 2 x^2+3 y^2=5\left(\frac{3 x+4 y}{5}\right)^2 \\ & \Rightarrow \quad 2 x^2+3 y^2=5\left(\frac{9 x^2+16 y^2+24 x y}{25}\right) \\ & \Rightarrow \quad 10 x^2+15 y^2=9 x^2+16 y^2+24 x y \\ & \Rightarrow \quad x^2-y^2+24 x y=0 \end{aligned}\) Here, (coefficient of \(\left.x^2\right)+\left(\right.\) coefficient of \(y^2\) ) \(\begin{aligned} & =1+(-1)=0 \\ \therefore \angle A O B ! & =90^{\circ} \text { (or) } \frac{\pi}{2} \end{aligned}\) \(\therefore\) Hence, answer is (c).

Asked in: AP EAMCET 2019 (23 Apr Shift 1)

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