दिये गए वक्र `x= a sin ^(3)theta` और `y=a cos ^(3)theta` के एक बिन्दु, जहाँ `theta = (pi)/(4)` है, पर
`(i)`स्पर्श रेखा का समीकरण ज्ञात कीजिए
`(ii)`अभिलम्ब का समीकरण ज्ञात कीजिए.

Options

A. `(i)x+y =(a)/(sqrt2)`
`(ii)x-y =1`
B. `(i)x+y =a`
`(ii)x-y =0`
C. `(i)x+y =(a)/(sqrt2)`
`(ii)x-y =0`
D. `(i)x-y =(a)/(sqrt2)`
`(ii)x+y =0`

Answer

C

Explanation

यहाँ `x=a sin ^(3)theta`
` implies(dx)/(d theta)=3a sin ^(2)theta cos theta`
और `y=a cos ^(3)theta`
`implies (dy)/(dx) =-3 a cos ^(2) theta sin theta`
`therefore (dy)/(dx) =(dy//d theta)/(dx//d theta)`
` (dy)/(dx) =(-3 a cos ^(2)theta sin theta)/(3 a sin ^(2)thetacos theta)`
`implies(dy)/(dx) =-cot theta`
`theta =(pi)/(4)` पर, `(dy)/(dx) =-cot ""(pi)/(4)=-1`
`implies [(dy)/(dx)]_((pi)/(4))=-1`
पुनः `theta =(pi)/(4)` पर `x=a (sin ""(pi)/(4))^(3) , y =a (cos ""(pi)/(4))^(2)`
` impliesx=a ((1)/(sqrt2))^(3), y =a ((1)/(sqrt2))^(3) ,`
`[because cos ""(pi)/(4)=sin ""(pi)/(4)=(1)/(sqrt2)]`
`implies x = (a)/((2)^(3//2)), y =(a)/((2)^(3//2))`
अब बिन्दु `((a)/((2)^(3//2)), (a)/(2^(3//2)))` पर स्पर्श रेखा का समीकरण है-
`(y- (a)/( 2 ^(3//2))=((dy)/(dx))_(theta =(pi)/(4))(x- (a)/(2 ^(3//2)))`
`impliesy- (a)/(2 ^(3//2))=(-1) (x-(a)/(2^(3//2)))`
`implies y-(a)/(2sqrt2)=(-1)(x- (a)/(2sqrt2))`
`implies y+x =(a)/(2sqrt2)+(a)/(2sqrt2)`
`impliesx+y=(2a)/(2sqrt2)`
` implies x+y =(a)/(sqrt2).`
बिन्दु `((a)/(2 ^(2//3)), (a)/(2 ^(3//2)))` पर अभिलम्ब का समीकरण है-
`y-(a)/(2 ^(3//2))=(-1)/(((dy)/(dx))_(theta =(pi)/(4)))(x- (a)/(2 ^(3//2)))`
`impliesy- (a)/(2 ^(3//2))=((-1))/((-1))(x- (a)/(2 ^(3//2)))`
`implies y-(a)/(2sqrt2)=1(x-(a)/(2sqrt2))`
`impliesy-(a)/(2sqrt2)=x- (a)/(2sqrt2)`
`impliesx-y =0.`