2x2+y2=20. . . (i) On differentiating w.r.t. x, we have
dy
dx
=
−2x
y
. . . (ii) and 4y2−x2=8 . . . (iii) Differentiating w.r.t. x, we have
dy
dx
=
x
4y
. . . (iv) Point of intersection of curve (i) and (ii) from Eq. (iii), x2=4y2−8 Putting in Eq. (i), 2(4y2−8)+y2=20 8y2−16+y2=20 9y2=36 y2=4 ⇒y=±2 ∴x2=4⋅(4)−8=8 or x=±2√2 ∴ Point of intersection in IVth quadrant (2√2,−2) Slope of tangent at (2√2,−2) of curve (1) m1=(
dy
dx
)(2√2,−2)=
−2×2√2
−2
=2√2 Slope of tangent at (2√2,−2) of curve (2) m2=(