(a) Tangent to the ellipe at `P(phi)` is
`(x)/(4) cos phi+(y)/(2) sin phi=1`
It must pass through the center of the circle. Hence,
`(4)/(4) cos phi+(2)/(2)sin phi=1`
`or cos phi+sin phi=1`
or `1+sin 2 phi phi=1`
or `sin 2phi=0`
i.e.,` 2phi=0 or pi`
i.e., `(phi)/(2)=0 or (pi)/(4)`
(b) Consider and point `P(sqrt(6) cos, theta, sqrt(2) sin theta)` on the ellipe `(x^(2))/(6)+(y^(2))/(2)=1`
Given thast P=2. Therefore, `6 cos^(2) theta+2sin^(2) theta=0`
or `4cos^(2) theta=2`
or `cos theta=+-(1)/(sqrt(2))`
i.e.,` theta=(pi)/(4) or (5pi)/(4)`
(c) Sovling the equation of ellipse and parbola (eliminating `(x^(2)`)n we have
`y-1+4y^(2)=4`
or `4y^(2)+y-5=0`
`(4y+5)(y-1)=0`
or y=1, x=0`
The curves touch at (0,1) so, the angle of intersection is 0.
(d) The normal at`P(a cos theta, b sin theta)` is
`(ax)/(cos thet)-(hx)/(sin theta)=a^(2)-b^(2)`
Where `a^(2)=14,b^(2)=5`
It means the cruve again at Q `(2 theta)`, i.e., `(a cos 2 theta, b sin 2theta)`
Hecen,
`(a)/(costheta)a cos 2theta-(b)/(sin theta)(b 2sintheta)=a^(2)-b^(2)`
`or (14)/(cos theta) cos 2theta-(5)/(sin theta)(sin 2 theta)=14-5`
or `28 cos^(2)theta-14-10 sec^(2)theta=9 cos theta`
`or 18 cos^(2)theta-9cos theta-14=0`
`or (6 cos theta-7)(3 cos theta+2)`
or `cos theta=-(2)/(3)`
