if the tangents from the point ( a , 3 ) to the ellipse      are at right angle, then a is

general equation of tangent of ellipse is

y = mx + (a²m² +b²)^1/2

for this ellipse we have

y = mx + (9m² +4)^1/2   --------------(1)

since (a,3) lies on this tangent

so, we have 3 = am + (9m² +4)^1/2

^{we can rewrite it as}

m²(a²-9) - 6am + 5 = 0

since tangents are at right angles so we have m₁m₂ = -1

5/(a² - 9) = -1

a² = 4  or a = 2

Any tangent to the ellipse can be written as xcosk+ysink=p=9cos²k+4sin²k.

this passes through (a,3)(acosk+3sink)²=9cos²k+4sin²k(a²-9)cot²k+6acotk+5=0.

since the tangents are perpendicular,cotk₁.cotk₂=-1=5/a²-9a=2or-2

well see another way is that .....



the locus of all the points from which the tangents to ellipse are mutually perpendicular is Director Circle of ellipse .



and equation of the director circle is {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}

since the point (a,3) are on the director circle ,

we have

{a}^{2} + 9 = 9 + 4

hence a=\pm2

well see another way is that .....



the locus of all the points from which the tangents to ellipse are mutually perpendicular is Director Circle of ellipse .



and equation of the director circle is {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}

since the point (a,3) are on the director circle ,

we have

{a}^{2} + 9 = 9 + 4

hence a=pm2