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- Find the point on the parabola that is closest to the point (0, -3). HINT: distance:=surd((x-x1)^2+9y-y1)^2,2); and parabola:=x+y^2=0; Isolate(x+y^2=0,x); Substitute the result obtained for x into your distance equation. Substitute the x and y ordinates (0 and -3) for x1 and y1 in your distance formula. Take the radicand (what’s under the radical of your new distance function) and differentiate that for y. Set this derivative to zero and solve for y. Substitute this y-value (select only the real number solution) into parabola equation to solve for x. You will now have both x and y ordinates.Of all the points on the graph of z = 10 - x^2 - y^2 that lie above the plane x + 2y + 3z = 0, what point is the farthest from the plane? (Do not use Langrange multipliers)What point on the parabola y = 1 - x^2 is closest to the point (1, 1)?
- What coordinate system is suggested if the integrand of a tripleintegral involves x2 + y2?VXEN, 3yEZ s.t. x + y^2 = x - y O TRUE FALSEThe coordinates of the points O, A and B of the plane are (0, 0), (0, 12) and (108, 0), respectively. A point P=(x,y) is such that the area of triangle ΔPOA is twice the area of triangle ΔPOB. We can then say that the coordinates of P satisfy the relationship: Choose an option: a. x^(2)−81/4y^(2)=0 b. x^(2)−324y^(2)=0 c. x^(2)−81y^(2)=0 d. x^(2)−18y^(2)=0 e.x^(2)−54y^(2)=0