TS EAMCET 5 May 2018 Shift 1 Solved Paper

The equation of parabola is,
y2=8x
Differentiate the above equation with respect to x.

dy

dx

=

4

y

At point P(2,4), the slope of tangent is,
m1=

4

4

=1
The equation of 1st tangent is,
y−4=1(x−2)
y=x+2 …… (1)
At point Q(18,−12), the slope of tangent is,
m2=4−12
=−

1

3

The equation of 2nd tangent is,
y+12=−

1

3

(x−18)
3y+18=−x …… (2)
From equation (1) and equation (2),
y=−4
x=−6
So, the tangents intersect at (-6,-4)
The required equation of tangent passes through (-6,-4) andhaving a slope

1

2

is,
y−(−4)=

1

2

[x−(−6)]
x−2y=2