If the line x -- 2y = 12 is tangent to the ellipse [x^2 a^2 + y^2 b^2 = 1] at the point [(…

Question

If the line x -- 2y = 12 is tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ at the point $$\left( {3, - {9 \over 2}} \right)$$ , then the length of the latus
rectum of the ellipse is :

Accepted Answer

Equation of tangent at $$\left( {3, - {9 \over 2}} \right)$$ to $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ is

$${{3x} \over {{a^2}}} - {{{y^9}} \over {2{b^2}}} = 1$$ which is equivalent to x -- 2y = 12

$${3 \over {{a^2}}} = {{ - 9} \over {2{b^2}( - 2)}} = {1 \over {12}}$$   (On comparing)

$${a^2} = 3 \times 12$$ and $${b^2} = {{9 \times 12} \over 4}$$

$$ \Rightarrow $$ a = 6 and b = $$3\sqrt 3 $$

So latus rectum = $${{2{b^2}} \over a} = {{2 \times 27} \over 6} = 9$$

Solutions · 1