Search any question & find its solution
Question:
Answered & Verified by Expert
The line $x+y=2$ is tangent to the curve $x^2=3-2 y$ at its point
Options:
Solution:
2054 Upvotes
Verified Answer
The correct answer is:
$(1,1)$
Given curve $x^2=3-2 y...(i)$
Differentiate w.r.t. $x, \quad 2 x=0-2 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-x$
Slope of the tangent of the curve $=-x$
From the given line, slope $=-1, \therefore x=1$ and from equation (i), $y=1$
$\therefore$ Co-ordinate of the point is $(1,1)$.
Differentiate w.r.t. $x, \quad 2 x=0-2 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-x$
Slope of the tangent of the curve $=-x$
From the given line, slope $=-1, \therefore x=1$ and from equation (i), $y=1$
$\therefore$ Co-ordinate of the point is $(1,1)$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.