Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

What is the minimum area of the triangle formed by any tangent to the ellipse x2a2+y2b2=1 with the coordinate axis.

Answer
VerifiedVerified
390.3k+ views
like imagedislike image
Hint: The standard form equation of the ellipse is given by x2a2+y2b2=1.The coordinates of any point lying on the ellipse is represented by (acosθ,bsinθ). The equation of the tangent at this point on the ellipse is given by xacosθ+ybsinθ=1.
To minimise the area of the triangle, we have to find the generalised form of the area formed using the tangent and ellipse and determine the condition to minimize the area.

Complete step-by-step solution:
The given ellipse has the equation x2a2+y2b2=1.
Choose any point R on the ellipse, so the coordinates of the point R are represented as R=(acosθ,bsinθ).
Let us draw the tangent at this point R on the ellipse, such that the tangent meets the y-axis at point Q and meets x-axis at point P.
The given figure represents the above ellipse and the mentioned tangent:
seo images

Now, the parametric form of the equation of the tangent at the point (acosθ,bsinθ) is given by xacosθ+ybsinθ=1.
The above equation can be written as
x(acosθ)+y(bsinθ)=1
So, the x and y coordinates of the above tangent equation are
P=(acosθ,0)
Q=(0,bsinθ)
Now, we get the three coordinates of the triangle formed by the tangent on the ellipse.
The triangle ΔQOPhas the coordinates,
P=(acosθ,0)Q=(0,bsinθ)O=(0,0)
Find the area of the above triangle, using the formulae: Area=12×base×height, where base is acosθ and height is bsinθ.
Area of ΔQOP=12×|acosθ×bsinθ|
Since, a and b cannot be negative so modulus can be removed from them,
Area of ΔQOP=ab×|12cosθsinθ|
Use the identity: 2sinθcosθ=sin2θ
Area of ΔQOP=ab×|1sin2θ|
To reduce the area of the triangle ΔQOP, we need to reduce the value of |1sin2θ|.
Since, |sin2θ|1
Taking the reciprocal, |1sin2θ|1
Since, the minimum value |1sin2θ|is 1, hence the minimum value of the area of ΔQOP is ab.
The minimum area of the triangle formed by any tangent to the ellipse x2a2+y2b2=1 with the coordinate axis is ab.

Note: The minimum value of sinθ can be 1 and the maximum value is 1. So, the value of sinθ always lies between 1 and 1, and can be represented as, 1<sinθ<1 or |sinθ|1.The equation of the form xa+yb=1 has x intercept a, and yintercept is b.