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Question

Find the equation of the tangents to the ellipse 4x2+9y2=72 which is perpendicular to the line 3x2y=5

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Solution

We will first find the slope of the tangent to the ellipse 4x2+9y2=72 using differentiation, as follows:
4x2+9y2=72
8x+18y(dydx)=0
(dydx)=4x9y

Now the slope of the given line 3x2y=5 is

y=(32)x52
slope of this line is 32
Now, this slope is perpendicular to the tangent of the ellipse, so we have:

4x9y32=1
4x9y=23
x=32y
Now , we will put this value of x in the equation of the ellipse to get the point at which the tangent is to be found.
4x2+9y2=72
4(32y)2+9y2=72
18y2=72
y=±2
x=32y=±3
We also have the slope of the tangent to ellipse as 23
Thus the equations of the tangents at (3,2) is :
y2=23(x3)
3y=2x+12
and another equation of the tangent at (3,2) is :
y+2=23(x+3)
3y=2x12

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