Question

Consider an ellipse `(x^2)/4+y^2=alpha(alpha` is parameter `>0)` and a parabola `y^2=8x` . If a common tangent to the ellipse and the parabola meets the coordinate axes at `Aa n dB` , respectively, then find the locus of the midpoint of `A Bdot`

Answer 1

The equation of tangent to `y^(2)=8x "at" (2t^(2),4t)` is `yt-x=2t^(2)=0" "(1)`
The equation of tangent to the ellipse `(x^(2))/(4alpha)+(y^(2))/(alpha)=1`
or `(2 sqrt(alpha) cos, theta, sqrt(alpha sin theta))` ltbrlt is ` (xcos theta)/(2sqrt(alpha))+(ysin theta)/(alpha))=1" "(2)`
Comparing (1)and (2), we get
`(sqrt(alpha))/(cos theta)=-t^(2),(sqrtalpha)/(sin theta)=2t" "(3)`
Let the midpoint of AB be (h,k).Them, `h=(sqrt(alpha))/(cos theta),k=(sqrt(alpha))/(2 sin theta)`
`:. h=-t^(2),k=t or k^(2)=-h`
or `y^(2)=-x" "` [From (3)]