If one of the roots of the equation $x^2-5 x-14=0$ is the length of the semi conjugate axis of the hyperbola…

If one of the roots of the equation $x^2-5 x-14=0$ is the length of the semi conjugate axis of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the square of the other root is the semi-transverse axis then the focus of the hyperbola that lies on the positive $x$-axis is
  1. $(5,0)$
  2. $(\sqrt{65}, 0)$
  3. $(7,0)$
  4. $(\sqrt{74}, 0)$

Solution

Given equation is $x^2-5 x-14=0$ how roots are $-2,7$ Let $b=7$ and $a=4$ hence $C=\sqrt{7^2+9^2}=\sqrt{65}$ hence $S=(\sqrt{65}, 0)$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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