4x2+9y2=36 Differrentiate borh sides with respect to x
8x+18yy′=0
y′=−9y4x
slope=m=−9y04x0
The equation of the tangent to the ellipse is
y−y0=m(x−x0)
y=mx−mx0+y0
x0=−49y0m
y0−mx0=4y0(4+9m2)
4x02+9y02=36
481y02m2+9y02=36
y02(9m2+4)=16
4y0=±9m2+41
y=mx±9m2+4
Point (−2,−3)
−3=m(−2)±9m2+4
2m−3=±9m2+4
4m2−12m+9=9m2+4
5m2+12m−5=0
m1m2=5−5=−1 Therefore the tangents from the point with position vector −2i−3j to the ellipse 4x2+9y2=36 are perpendicular.
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