If y=x+c touches the ellipse 3x^2+4y^2=12 at the point P, then the value of the length OP (where O is the origin) is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) 5/(√7) units. If y=x+c touches (x^2)/4+(y^2)/3=1 then, c=±√(41+3) =±√7 Let's assume, c=√7 and y=x+√7 touches (x^2)/4+(y^2)/3=1 at h,k. then, xh/4+yk/3=1 is same as y-x=√7 ⇒h/4-1=k/31=1/(√7) ⇒h,k=-4/(√7),3/(√7)=P Hence, OP=√(167+97)=5/(√7) units

If $y = x + c$ touches the ellipse $3 x^{2} + 4 y^{2} = 12$ at the point $P,$ then the value of the length $O P$ (where $O$ is the origin) is equal to

Solution

More Ellipse Questions

If y=x+c touches the ellipse 3x2+4y2=12 at the point P, then the value of the length OP (where O is the origin) is equal to

Solution

More Ellipse Questions

If $y = x + c$ touches the ellipse $3 x^{2} + 4 y^{2} = 12$ at the point $P,$ then the value of the length $O P$ (where $O$ is the origin) is equal to