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Question: Find the equation of tangent to the ellipse \[4{{x}^{2}}+9{{y}^{2}}=72\] which is perpendicular to t...

Find the equation of tangent to the ellipse 4x2+9y2=724{{x}^{2}}+9{{y}^{2}}=72 which is perpendicular to the line 3x−2y=53x-2y=5

Explanation

Solution

Hint: First we have to convert the given ellipse equation to the general form and obtain the values ofa2{{a}^{2}}and b2{{b}^{2}}. We have know that if two lines are perpendicular then the product of their slopes is equal to −1-1

Complete step-by-step answer:

Given the equation of ellipse is 4x2+9y2=724{{x}^{2}}+9{{y}^{2}}=72 and we have to find the tangent which is perpendicular to the line 3x−2y=53x-2y=5
Firstly convert the given ellipse equation to general form that is x2a2+y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1
So by converting the given equation to the general form we will get equation as follows
x218+y28=1\dfrac{{{x}^{2}}}{18}+\dfrac{{{y}^{2}}}{8}=1
So by comparing the obtained equation with general form we will get the following values,
a2=18{{a}^{2}}=18
b2=8{{b}^{2}}=8
We know that the equation of tangent to the ellipse in slope form is
y=mx+a2m2+b2y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}
But in the question given that tangent is perpendicular to the line 3x−2y=53x-2y=5
So slope of the line 3x−2y=53x-2y=5is 32\dfrac{3}{2}
We know that when two lines are perpendicular to each other then m1×m2=−1\mathop{m}_{1}\times \mathop{m}_{2}=-1
32×m2=−1\dfrac{3}{2}\times \mathop{m}_{2}=-1
m2=−23\mathop{m}_{2}=\dfrac{-2}{3}
So the slope of the tangent to the given ellipse equation is −23\dfrac{-2}{3}
Now substitute m=−23m=\dfrac{-2}{3} in the equation of tangent to the ellipse in slope form we will get
y=mx+a2m2+b2y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}
Now substitute the obtained values in the given expression we will get,
y=−23x+18(49)+8y=\dfrac{-2}{3}x+\sqrt{18\left( \dfrac{4}{9} \right)+8}
y=−23x+16y=\dfrac{-2}{3}x+\sqrt{16}
One equation of tangent to the ellipse is
y=−23x+4y=\dfrac{-2}{3}x+4
2x+3y−12=02x+3y-12=0
Other equation of tangent to the ellipse is
y=−23x−4y=\dfrac{-2}{3}x-4
2x+3y+12=02x+3y+12=0
Hence we get the required equation of tangents to the given ellipse
Note: We have to note that general equation of tangents to any given ellipse is of the form y=mx+a2m2+b2y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}and in the general form of the ellipsex2a2+y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 if a2{{a}^{2}} is greater than b2{{b}^{2}}then a is called as major axis and b is called as minor axis in ellipse