Answer to Question #105062 in Analytic Geometry for Fifty Adawu

Question

Show that the tangent from the point with vector -2i-3j to the ellipse 4x^2 +9y^2=36 are perpendicular?

Answer 1

$ellipse :4x^2 +9y^2=36$

Converting this into standard form

it becomes : $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$

$a^2=9 \ \&\ b^2=4$

Given vector : $-2i-3j$

Converting this into point it becomes (-2,-3)

Let the tangent to the ellipse from point $y=mx+c$

then it must satisfy the condition $c^2=a^2m^2+b^2$

substituting the values in both the equation we get :

$-3=-2m+c$ ……..(1) & $c^2=9m^2+4$ ……..(2)

Putting the value of c in the eq(2) from the equation 1

$(2m-3)^2=9m^2+4\\4m^2+9-12m=9m^2+4\\5m^2+12m-5=0\\$

From here we will get two values of m(slope)

If the tangents are perpendicular then the product of slopes must be -1

product of roots of the above quadratic = -5/5=-1