QJEEAdv16Mad140: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Mad140 Solution: Let P be the point on the parabola y 2 = 4 x which is at the…

Tangent to y² = 4x at (t², 2t) is

y(2t) = 2(x + t²)

$$\Rightarrow$$ yt = x + t² ...... (i)

Equation of normal at P(t², 2t) is

y + tx = 2t + t³

Since, normal at P passes through centre of circle S(2, 8).

$$\therefore$$ 8 + 2t = 2t + t³ $$\Rightarrow$$ t = 2, i.e., P(4, 4)

$$\therefore$$ $$SP=\sqrt{{(4-2)}^2+{(4-8)}^2}=2\sqrt{5}$$

$$\therefore$$ Option (a) is correct.

Also, SQ = 2

$$\therefore$$ PQ = SP $$-$$ SQ = 2$$\sqrt{5}$$ $$-$$ 2

Thus, $$\frac{SQ}{QP}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$$

$$\therefore$$ Option (b) is incorrect.

Now, x-intercept of normal is

x = 2 + 2² = 6

$$\therefore$$ Option (c) is correct.

Slope of tangent $$=\frac{1}{t}=\frac{1}{2}$$

$$\therefore$$ Option (d) is correct.