JEE Main 26-Aug-2021 Shift 1 Solved Paper

Equation of tangent to y2=30x is,
y=mx+

30

4m

Now, this tangent passes through (−30,0).
∴0=−30m+

30

4m

⇒

30

4m

=30m
⇒ m2=

1

4

⇒ m=±

1

2

∴ Equation of tangent is
y=

x

2

+15
ory=−

x

2

−15
Now equation of circle is x2+y2+30x+

675

4

=0
Let perpendicular distance of the tangent from the centre (−15,0) of the circle = p
∴ p=

| −15 2 +15|

√1+ 1 4

=3√5
∴ Length of chord =2√r2−p2
=2√(152+0−

675

4

)−45=3√5
where, r is radius of the given circle.