1.Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2−9x2=116y^2 - 9x^2 = 116y2−9x2=1 isa.x2+y2=9x^2 + y^2 = 9x2+y2=9b.x2+y2=19x^2 + y^2 = \frac{1}{9}x2+y2=91c.x2+y2=7144x^2 + y^2 = \frac{7}{144}x2+y2=1447d.x2+y2=116x^2 + y^2 = \frac{1}{16}x2+y2=161Login to continueOnly logged in users canattempt or see the solution.
