# Please answer immediately. If the normal at the point P(thetha) to the ellipse x2/14+y2/5=1 intersect it again at the point Q(2thetha) then cos(thetha) is??

Saurabh Koranglekar
3 years ago
Vikas TU
14149 Points
3 years ago
Normal at the point P(theta) to the ellipse x²/14 + y²/5 = 1 intersects it again at the point Q(2theta).
we know, standard equation of ellipse is
x²/a² + y²/b² = 1 compare it with given equation
so, a² = 14 then, a = √14
b² = 5 then, b = √5
now equation of normal passing through point P(theta) is given by,
ax/cos theta - by/sin theta = a² - b².
or, √14x/cos theta - √5y/sin theta = 14 - 5 = 9 ....(1)
it again meets the curve at the point Q(2theta)
so, Q(2theta) = (√14cos2theta, √5sin2theta)
now, put it in equation (1),
or, 14cos2theta/cos theta - 5sin2theta/sintheta = 9
or, 14(2cos² theta - 1)/cos theta - 10sin theta cos theta/sin theta = 9
or, 28cos theta - 14sec theta - 10cos theta = 9
or, 18cos theta - 14/cos theta = 9
or, 18cos²theta - 14 - 9cos theta = 0
or, 18cos²theta -21cos theta + 12costheta - 14 = 0
or, 3cos theta(6cos theta - 7) + 2(cos theta - 7) = 0
or, (3cos theta + 2)(6cos theta - 7) = 0
or, cos theta = -2/3 [ because cos theta ≠ 7/6 ]