Question

Show that the tangents from the point with position vector −2i − 3j to the ellipse
4x
2 + 9y
2 = 36 are perpendicular.

Answer 1

Equation of ellipse is given as

$4x^2 + 9y^2 = 36$ , the general form of equation is

$\frac{x^2}{9} +\frac{y^2}{4} = 1$

on comparing with general equation of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1$

here, a= 3 , and b=2,

here the position vector of point is given as -2i-3j so the coordinate of point will be (-2,-3)

Now, we know that equation of tangent to the given ellipse with m as slope will be

$y= mx+\sqrt{a^2m^2+ b^2}$

$y= mx+ \sqrt{9m^2+4}$

and it passes through the point (-2,-3)

so the equation become

$-3=-2m+\sqrt{9m^2+4}$

$2m-3=\sqrt{9m^2+4}$

now squaring the both sides we get

$(2m-3)^2= (\sqrt{9m^2+4})^2$

$4m^2+9-12m= 9m^2+4$

$5m^2+12m-5=0$

this is quadratic equation we can find the value of m by solving this equation using quadratic formula,

$m_1=\frac{-12 +\sqrt{144+100}}{10} ,m_2=\frac{-12 -\sqrt{144+100}}{10}$ .

These are the slopes of two tangents from the ellipse passing through the point(-2,-3),

now both will be perpendicular is

$m_1m_2=-1$

LHS$=$ $\frac{-12 +\sqrt{144+100}}{10} \times \frac{-12 -\sqrt{144+100}}{10}$

LHS=-1 = RHS

so the tangents are perpendicular

Hence proved.........