Let the parabola y=−x2−2x+k and the parabola y=−
1
2
x2−4x+3 touches the point P(x1,y1) Now, y=−x2−2x+k
dy
dx
)(x1,y1)=−2x1−2 .......(i) and y=−
1
2
x2−4x+3 (
dy
dx
)=−x−4 (
dy
dx
)(x1,y1)=x1−4 .......(ii) Since, parabola touches of (x1,y1). ∴ Slope of their tangents are equal −2x1−2=−x1−4 ∴x1=2 put the value of x1 in y1=−