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

If β is one of the angles between the normals to the ellipse, x2+3y2=9 at the points (3cosθ,3sinθ) and (3sinθ,3cosθ) , θ(0,π2) then 2cotβsin2θ is equal to
A. 2
B. 23
C. 13
D. 34

Answer
VerifiedVerified
471.9k+ views
like imagedislike image
Hint: We find the expression for slope of the tangent at any point by differentiating the given curve and then the slope of the normal using perpendicular condition. We find the slopes of the intersecting normals as m1,m2 by putting given points (3cosθ,3sinθ) and (3sinθ,3cosθ) in the slope expression. We use the formula for angle between two lines with slopes tanβ=|m1m21+m1m2| to find 2cotβsin2θ .

Complete step by step answer:
We know from differential calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point.
We are given equation of the ellipse as
 x2+3y2=9x2+3y29=0
Let us differentiate equation of above ellipse with respect to x and find the expressions for slope at any point .We have
ddx(x2+3y29)=02x+6ydydx=0dydx=x3y
We know that the product of slopes of two perpendicular line is 1 . So we can find the slope expression for normal as
1dydx=1x3y=3yx
Since we are given there are two normals with points of contact points (3cosθ,3sinθ) and (3sinθ,3cosθ) . Let us denote their slopes as m1 and m2 respectively. Let us find the slopes of the normal by putting the points in expression for slope of normal. We have;
m1=33sinθ3cosθ=3tanθm2=33cosθ3sinθ=3cotθ
We are given that the angle between the normals is β . We use the formula for angle between two lines and have;
tanβ=|m1m21+m1m2|tanβ=|3tanθ(3cotθ)1+3tanθ(3)cotθ|tanβ=|3(tanθ+cotθ)13tanθcotθ|
Since tangent and co-tangent are reciprocal to each other we have;
1cotβ=|3(tanθ+cotθ)13|1cotβ=|3(tanθ+cotθ)2|1cotβ=32|tanθ+cosθ|
We convert the tangent and co-tangents into sine and cosine to have;
1cotβ=32|sinθcosθ+cosθsinθ|1cotβ=32|sin2θ+cos2θsinθcosθ|1cotβ=3|sin2θ+cos2θ2sinθcosθ|
We use the Pythagorean trigonometric identity sin2θ+cos2θ=1 and sine double angle formula sin2θ=2sinθcosθ to have;
1cotβ=3|1sin2θ|1cotβ=31sin2θcotβsin2θ=13
Hence the correct option is C.

seo images

Note:
We note that since we are given θ(0,π2) we have 2θ(0,π) . Since sine is positive for all values in the interval (0,π) we have |sin2θ|=sin2θ . We note that the equation of tangent to the standard ellipse x2a2+y2b2=1 is y=mx±a2m2+b2 . We also note that tanβ=|m1m21+m1m2| gives us only one angle between the intersecting lines, the other angle is πβ .