Question

Find where the line $2x+y=3$ cuts the curve $4x^2+y^2=5.$ Obtain the equations of the normals at the points of intersection and determine the co-ordinates of the point where these normals cut each other.

• A. $(-1,\frac{1}{2})$
• B. $(1,\frac{1}{2})$
• C. $(-1,\frac{-1}{2})$
• D. $(1,\frac{-1}{2})$

Answer 1

The correct option is A $(-1,\frac{1}{2})$

$P(\frac{1}{2}+,2),Q(1,1)$
Tangents at P and Q are $4x\cdot \frac{1}{2}+y\cdot 2=5$ and $4x\cdot 1+y\cdot 1=5$ or $2x+2y=5$ and $4x+y=5$
Hence normals are $2x-2y+3=0,$ $x-4y+3=0$
They intersect at $(-1,\frac{1}{2})$