The coordinates of the point $$P(x$$,$$y)$$ lying in the first quadrant on the ellipse $$x2/8+y2/18=1$$ so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, are given by

Question

The coordinates of the point $$P(x$$,$$y)$$ lying in the first quadrant on the  ellipse $$x2/8+y2/18=1$$ so that the area of the triangle formed by the  tangent at P and the coordinate axes is the smallest, are given by

Infinity Learn · Accepted Answer

Any point on the ellipse is given by (8cos$$⁡θ,$$18sin$$⁡θ$$) Now  $$2x8+218ydydx=0$$⇒$$dydx=$$−$$9x4y$$⇒$$dydx(8cos$$⁡θ,$$18sin$$⁡θ$$)=$$−$$98cos$$⁡θ$$418sin$$⁡θ$$=$$     −$$92cot$$θ Hence the equation of the tangent at (8cos$$⁡θ,$$18sin$$⁡θ$$) is y   −$$18$$ $$sin$$θ  $$=$$  $$92$$ $$cot$$θ (x$$−$$8$$  $$cos$$θ$$)Therefore$$, the tangent cuts the coordinate axes at the $$points0$$,  $$18sin$$θ    and    $$8cos$$θ, $$0Thus$$ the area of the triangle formed by this tangent and the  coordinate axes is $$A=1218$$⋅$$8$$⋅$$1cos$$⁡θ$$sin$$⁡θ$$=6cos$$⁡θ$$sin$$⁡θ$$=12cosec$$⁡$$2$$θ But cosec $$2$$θ is smallest when θ$$=$$π$$/4$$. Therefore A is  smallest when θ$$=$$π$$/4Hence$$ the required point is $$8$$⋅$$12$$⋅$$18$$⋅$$12=(2$$,$$3)

The coordinates of the point P(x,y) lying in the first quadrant on the ellipse x2/8+y2/18=1 so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, are given by

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!