TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The equation of a tangent to y2=4√kx is
y=mx+√k/m, where m is the slope of tangent.
If it touches the 2x2+2y2=k, then
⇒m√1+m2=√2⇒m4+m2−2=0
⇒(m2+2)(m2−1)=0⇒m=±1
Substituting these values in y=mx+‌

√k

m

, the equation of common tangents are y=x+√k and
y=−x−√k
∴‌‌m1=1,m2=−1
Now, m1m2=1×−1=−1