Question

how many times line y = 2x meets the graph of y = sec2tan-1 x​

Answer 1

Step-by-step explanation:

Answer

Correct option is

D

2

3

5

x=sec

2

t

y=cott

dx

dy

=−

2sec

2

t⋅tant

csc

2

t

=−

2

cot

3

t

Now,

dx

dy

∣

∣

∣

∣

∣

t=45

0

=−

2

1

x∣

t=45

0

=2

y∣

t=45

0

=1

Hence the equation of the tangent will be

y−1=−

2

1

(x−2)⇒2y−2=−x+2⇒x+2y=4 ...(i)

Now,

sec

2

t=1+tan

2

t

sec

2

t=1+

cot

2

t

1

Or

x=1+

y

2

1

⇒xy

2

=y

2

+1 is the equation of curve.

Solving the equation of tangent and equation of curve we get

4−2y=1+

y

2

1

Or

4y

2

−2y

3

=y

2

+1⇒2y

3

−3y

2

+1=0

y=−

2

1

and y=1

For y=1;x=2

For y=−

2

1

;x=5

Hence,

Q=(5,−

2

1

) and P=(2,1)

∴PQ=

2

3

5