Question #224017

Show that the tangents from the point with position vector −2i − 3j to the ellipse 4x2 + 9y2 = 36 are perpendicular.


1
Expert's answer
2021-08-16T15:17:46-0400
4x2+9y2=364x^2+9y^2=36

Differrentiate borh sides with respect to xx


8x+18yy=08x+18yy'=0

y=4x9yy'=-\dfrac{4x}{9y}

slope=m=4x09y0slope=m=-\dfrac{4x_0}{9y_0}

The equation of the tangent to the ellipse is


yy0=m(xx0)y-y_0=m(x-x_0)

y=mxmx0+y0y=mx-mx_0+y_0

x0=9y04mx_0=-\dfrac{9y_0}{4}m

y0mx0=y0(4+9m2)4y_0-mx_0=\dfrac{y_0(4+9m^2)}{4}

4x02+9y02=364x_0^2+9y_0^2=36

81y024m2+9y02=36\dfrac{81y_0^2}{4}m^2+9y_0^2=36

y02(9m2+4)=16y_0^2(9m^2+4)=16

y04=±19m2+4\dfrac{y_0}{4}=\pm\dfrac{1}{\sqrt{9m^2+4}}

y=mx±9m2+4y=mx\pm\sqrt{9m^2+4}


Point (2,3)(-2, -3)


3=m(2)±9m2+4-3=m(-2)\pm\sqrt{9m^2+4}

2m3=±9m2+42m-3=\pm\sqrt{9m^2+4}

4m212m+9=9m2+44m^2-12m+9=9m^2+4

5m2+12m5=05m^2+12m-5=0

m1m2=55=1m_1m_2=\dfrac{-5}{5}=-1

Therefore the tangents from the point with position vector 2i3j-2i-3j to the ellipse 4x2+9y2=364x^2+9y^2=36 are perpendicular.


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