Let x2/3+y2/3=a2/3,(a>0) be the equation of curve then which of the following is/are correct?
A
Equation of tangent at (x1,y1) on the curve is xx1/31+yy1/31=a2/3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Length of portion of tangent intercepted between the coordinate axes is constant.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
If the curve x2/3+y2/3=a2/3 touches the curve x2c2+y2d2=1 then c+d=a
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Equation of normal at (x1,y1) on the curve is x.x1/31−y.y1/31=x4/31+y4/31
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are A Equation of tangent at (x1,y1) on the curve is xx1/31+yy1/31=a2/3 B Length of portion of tangent intercepted between the coordinate axes is constant. C If the curve x2/3+y2/3=a2/3 touches the curve x2c2+y2d2=1 then c+d=a (x,y)=(acos3θ,asin3θ) dydx=−sinθcosθ Equation of tangent →(y−asin3θ)=−sinθcosθ(x−acos3θ) →xcosθ+ysinθ=a →xx1/31+yy1/31=a2/3
Length of the portion of the tangent between the coodinate axes : AB=√a2cos2θ+a2sin2θ=a
Tangent at (x1y1)tox2c2+y2d2=1 is xx1c2+yy1d2=1 which is identical to xx1/31+xy1/31=a2/3 So x1c2.×x1/31=y1cl2×y1/31=1a2/3 x4/31=c2a2/3,y4/31=d2a2/3 x2/31=ca1/3,y2/31=da2/3
which lies on x2/3+y2/3=a2/3 ca1/3+da1/3=a2/3 c+d=a
Equation of normal y′=y1/3x1/3 So normal is (y−y1)=x1/31y1/3(x−x1)