Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + px + p^3 = 0$ $(p \neq 0)$. If $(\alpha, \beta)$ is a point on the parabola $y^2 = x$,then the roots of the quadratic equation are:

  • A
    $4, -2$
  • B
    $-4, -2$
  • C
    $4, 2$
  • D
    $-4, 2$

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Let $F_1(-1, 0)$ and $F_2(1, 0)$ be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having its vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
$(1)$ The orthocentre of the triangle $F_1 M N$ is
$(A)$ $\left(-\frac{9}{10}, 0\right)$ $(B)$ $\left(\frac{2}{3}, 0\right)$ $(C)$ $\left(\frac{9}{10}, 0\right)$ $(D)$ $\left(\frac{2}{3}, \sqrt{6}\right)$
$(2)$ If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$,then the ratio of the area of the triangle $M Q R$ to the area of the quadrilateral $M F_1 N F_2$ is
$(A)$ $3: 4$ $(B)$ $4: 5$ $(C)$ $5: 8$ $(D)$ $2: 3$

If the normal to the rectangular hyperbola $xy = c^2$ at the point $t$ meets the curve again at $t_1$,then $t^3 t_1$ has the value equal to

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If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

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