TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

We have, x−2y+k=0 is tangent to the parabola y2−4x−4y+8=0, then
Now put x=2y−k in equation of parabola, we get
∴‌‌y2−4(2y−k)−4y+8=0
⇒‌‌y2−8y+4k−4y+8=0
⇒‌‌y2−12y+4k+8=0
Since line is tangent to parabola
∴‌‌D=0
(−12)2−4(4k+8)=0
⇒144−16k−32=0
⇒16k=112
⇒k=7
Now, we have
y2−4x−4y+8=0
∴‌‌2y‌

dy

dx

−4−4‌

dy

dx

=0
⇒‌‌2‌

dy

dx

(y−2)=4
⇒‌‌‌

dy

dx

=‌

2

y−2

∴ Slope of tangent at (l,k)=‌

2

k−2

=‌

2

7−2

‌‌[∵k=7]
=‌

2

5