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Match the following lists: |
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Answer» (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)` |
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