Question

A planet follows an elliptical path described by 16 x squared + 4 y squared = 64. A comet follows the parabolic path y equals x squared minus 4. Where might the comet intersect the orbiting​ planet?

Answer 1

Answer:

The comet might intersect the orbiting planet at (0, -4),$(\pm\frac{\sqrt{31}}{2},\frac{15}{4})$.

Step-by-step explanation:

Given that, a planet follows an elliptical path described by

$16x^2+4y^2=64$ .........(1)

A comet follows the parabolic path

$y=x^2-4$

$x^2=y+4$.........(2)

To find the intersecting point, we need to solve the above equations.

Putting  $x^2=y+4$  in the equation (1)

$y+4+4y^2=64$

$\Rightarrow 4y^2+y+4-64=0$

$\Rightarrow 4y^2+y-60=0$

$\Rightarrow 4y^2+16y-15y-60=0$

$\Rightarrow 4y(y+4)-15(y+4)=0$

$\Rightarrow(y+4)(4y-15)=0$

$\Rightarrow y=-4, \frac{15}{4}$

When y = -4 , then $x^2=-4+4$ $\Rightarrow x=0$

When $y=\frac{15}{4}$, then $x^2=\frac{15}{4}+4$ $\Rightarrow x=\pm \frac{\sqrt{31}}{2}$

The comet might intersect the orbiting planet at (0, -4),$(\pm\frac{\sqrt{31}}{2},\frac{15}{4})$.